ChemistryQ151–250 of 716 questions
Page 4 of 8 · English
151
ChemistryEasyMCQMHT CET · 2023
What is the value of $\angle S-S-S$ in the puckered $S_8$ ring of rhombic sulfur (in $^{\circ}$)?
A
$107$
B
$120$
C
$104.5$
D
$60$
Solution
(A) The $S_8$ molecule has a puckered ring structure, often described as a crown-shaped conformation.
In this structure, each sulfur atom is bonded to two other sulfur atoms with a bond angle of $107^{\circ}$ and a bond length of $204 \text{ pm}$.
In this structure, each sulfur atom is bonded to two other sulfur atoms with a bond angle of $107^{\circ}$ and a bond length of $204 \text{ pm}$.
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152
ChemistryEasyMCQMHT CET · 2023
Identify the elements undergoing reduction and oxidation respectively in the following redox reaction.
$3 H_3AsO_{3(aq)} + BrO_{3(aq)}^{-} \rightarrow Br_{(aq)}^{-} + 3 H_3AsO_{4(aq)}$
$3 H_3AsO_{3(aq)} + BrO_{3(aq)}^{-} \rightarrow Br_{(aq)}^{-} + 3 H_3AsO_{4(aq)}$
A
$As$ and $O$
B
$Br$ and $As$
C
$As$ and $Br$
D
$Br$ and $O$
Solution
(B) In the reaction $3 H_3AsO_3 + BrO_3^- \rightarrow Br^- + 3 H_3AsO_4$,we assign oxidation numbers:
For $BrO_3^-$,the oxidation state of $Br$ is $+5$. In $Br^-$,it is $-1$. Since the oxidation number decreases from $+5$ to $-1$,$Br$ undergoes reduction.
For $H_3AsO_3$,the oxidation state of $As$ is $+3$. In $H_3AsO_4$,it is $+5$. Since the oxidation number increases from $+3$ to $+5$,$As$ undergoes oxidation.
Therefore,$Br$ undergoes reduction and $As$ undergoes oxidation.
For $BrO_3^-$,the oxidation state of $Br$ is $+5$. In $Br^-$,it is $-1$. Since the oxidation number decreases from $+5$ to $-1$,$Br$ undergoes reduction.
For $H_3AsO_3$,the oxidation state of $As$ is $+3$. In $H_3AsO_4$,it is $+5$. Since the oxidation number increases from $+3$ to $+5$,$As$ undergoes oxidation.
Therefore,$Br$ undergoes reduction and $As$ undergoes oxidation.
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153
ChemistryEasyMCQMHT CET · 2023
Which of the following elements exhibits the usual tendency to undergo reduction?
A
$Mg$
B
$Ni$
C
$O$
D
$Cu$
Solution
(C) Metals,such as $Mg$,$Ni$,and $Cu$,generally have a low ionization energy and exhibit a tendency to lose electrons,which is known as oxidation.
Non-metals,such as $O$,have a high electronegativity and a strong tendency to gain electrons,which is known as reduction.
Non-metals,such as $O$,have a high electronegativity and a strong tendency to gain electrons,which is known as reduction.
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154
ChemistryMediumMCQMHT CET · 2023
What is the number of moles of electrons gained by one mole of oxidizing agent in the following redox reaction?
$Zn_{(s)} + 2HCl_{(aq)} \longrightarrow ZnCl_2(aq) + H_2(g)$
$Zn_{(s)} + 2HCl_{(aq)} \longrightarrow ZnCl_2(aq) + H_2(g)$
A
$4$
B
$3$
C
$2$
D
$1$
Solution
(D) In the given reaction: $Zn_{(s)} + 2HCl_{(aq)} \longrightarrow ZnCl_2(aq) + H_2(g)$
$1$. The oxidation state of $Zn$ changes from $0$ to $+2$ (loss of $2$ electrons,oxidation).
$2$. The oxidation state of $H$ in $HCl$ changes from $+1$ to $0$ in $H_2$ (gain of electrons,reduction).
$3$. Since $H^+$ is being reduced,$HCl$ acts as the oxidizing agent.
$4$. The reduction half-reaction is: $2H^+ + 2e^- \longrightarrow H_2$.
$5$. This shows that $2$ moles of $H^+$ (which corresponds to $2$ moles of $HCl$) gain $2$ moles of electrons.
$6$. Therefore,$1$ mole of the oxidizing agent $(HCl)$ gains $1$ mole of electrons.
$1$. The oxidation state of $Zn$ changes from $0$ to $+2$ (loss of $2$ electrons,oxidation).
$2$. The oxidation state of $H$ in $HCl$ changes from $+1$ to $0$ in $H_2$ (gain of electrons,reduction).
$3$. Since $H^+$ is being reduced,$HCl$ acts as the oxidizing agent.
$4$. The reduction half-reaction is: $2H^+ + 2e^- \longrightarrow H_2$.
$5$. This shows that $2$ moles of $H^+$ (which corresponds to $2$ moles of $HCl$) gain $2$ moles of electrons.
$6$. Therefore,$1$ mole of the oxidizing agent $(HCl)$ gains $1$ mole of electrons.
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155
ChemistryEasyMCQMHT CET · 2023
Which element from the following rapidly loses its luster in air and tarnishes?
A
$Ba$
B
$Be$
C
$K$
D
$Mg$
Solution
(C) Elements of group $1$ (alkali metals) are highly reactive. They rapidly lose their metallic luster in air due to the formation of a surface layer of oxide,peroxide,or superoxide by reacting with oxygen and moisture in the air. Among the given options,$K$ (potassium) is an alkali metal belonging to group $1$,while $Ba$,$Be$,and $Mg$ are alkaline earth metals (group $2$),which are relatively less reactive.
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156
ChemistryMediumMCQMHT CET · 2023
What is the oxidation state of carbon in $CaC_2$ and $K_2C_2O_4$ respectively?
A
$-2$ and $+6$
B
$-1$ and $+3$
C
$+2$ and $+2$
D
$-2$ and $+3$
Solution
(B) For $CaC_2$: The oxidation state of $Ca$ is $+2$. Let the oxidation state of $C$ be $x$. Thus,$+2 + 2x = 0$,which gives $x = -1$.
For $K_2C_2O_4$: The oxidation state of $K$ is $+1$ and $O$ is $-2$. Let the oxidation state of $C$ be $x$. Thus,$2(+1) + 2x + 4(-2) = 0$,which simplifies to $2 + 2x - 8 = 0$,so $2x = 6$ or $x = +3$.
Therefore,the oxidation states are $-1$ and $+3$ respectively.
For $K_2C_2O_4$: The oxidation state of $K$ is $+1$ and $O$ is $-2$. Let the oxidation state of $C$ be $x$. Thus,$2(+1) + 2x + 4(-2) = 0$,which simplifies to $2 + 2x - 8 = 0$,so $2x = 6$ or $x = +3$.
Therefore,the oxidation states are $-1$ and $+3$ respectively.
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157
ChemistryEasyMCQMHT CET · 2023
What is the stock notation of Manganese dioxide?
A
$Mn(I)O_2$
B
$Mn(II)O_2$
C
$Mn(III)O_2$
D
$Mn(IV)O_2$
Solution
(D) The chemical formula for manganese dioxide is $MnO_2$.
To find the oxidation state of $Mn$,let it be $x$.
The oxidation state of oxygen is $-2$.
So,$x + 2(-2) = 0$.
$x - 4 = 0$,which gives $x = +4$.
Therefore,the stock notation is $Mn(IV)O_2$.
To find the oxidation state of $Mn$,let it be $x$.
The oxidation state of oxygen is $-2$.
So,$x + 2(-2) = 0$.
$x - 4 = 0$,which gives $x = +4$.
Therefore,the stock notation is $Mn(IV)O_2$.
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158
ChemistryEasyMCQMHT CET · 2023
Oxidation state of manganese in potassium permanganate is:
A
$+3$
B
$+7$
C
$+5$
D
$+1$
Solution
(B) The chemical formula for potassium permanganate is $KMnO_4$.
Oxidation number of $K = +1$.
Oxidation number of $O = -2$.
Since $KMnO_4$ is a neutral molecule,the sum of the oxidation numbers of all atoms must be $0$.
Let the oxidation number of $Mn$ be $x$.
$(+1) + x + 4 \times (-2) = 0$.
$1 + x - 8 = 0$.
$x - 7 = 0$.
$\therefore x = +7$.
Thus,the oxidation state of manganese in $KMnO_4$ is $+7$.
Oxidation number of $K = +1$.
Oxidation number of $O = -2$.
Since $KMnO_4$ is a neutral molecule,the sum of the oxidation numbers of all atoms must be $0$.
Let the oxidation number of $Mn$ be $x$.
$(+1) + x + 4 \times (-2) = 0$.
$1 + x - 8 = 0$.
$x - 7 = 0$.
$\therefore x = +7$.
Thus,the oxidation state of manganese in $KMnO_4$ is $+7$.
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159
ChemistryEasyMCQMHT CET · 2023
What is the oxidation number of sulfur in $H_2SO_5$?
A
$+4$
B
$+6$
C
$+8$
D
$+5$
Solution
(B) $H_2SO_5$ is Peroxymonosulfuric acid (Caro's acid).
It contains one peroxide linkage $(-O-O-)$.
Let the oxidation number of sulfur be $x$.
The oxidation number of $H$ is $+1$,and for oxygen,there are three atoms with $-2$ and two atoms in the peroxide linkage with $-1$.
$(2 \times (+1)) + x + (3 \times (-2)) + (2 \times (-1)) = 0$
$2 + x - 6 - 2 = 0$
$x - 6 = 0$
$x = +6$
It contains one peroxide linkage $(-O-O-)$.
Let the oxidation number of sulfur be $x$.
The oxidation number of $H$ is $+1$,and for oxygen,there are three atoms with $-2$ and two atoms in the peroxide linkage with $-1$.
$(2 \times (+1)) + x + (3 \times (-2)) + (2 \times (-1)) = 0$
$2 + x - 6 - 2 = 0$
$x - 6 = 0$
$x = +6$
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160
ChemistryEasyMCQMHT CET · 2023
Which among the following compounds exhibits $+2$ oxidation state of oxygen?
A
$H_2O$
B
$SO_2$
C
$OF_2$
D
$H_2O_2$
Solution
(C) The oxidation number of $F$ is $-1$ in all of its compounds.
Let the oxidation state of $O$ be $x$.
In $OF_2$,the sum of oxidation states is $x + 2(-1) = 0$.
Therefore,$x - 2 = 0$,which gives $x = +2$.
Let the oxidation state of $O$ be $x$.
In $OF_2$,the sum of oxidation states is $x + 2(-1) = 0$.
Therefore,$x - 2 = 0$,which gives $x = +2$.
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161
ChemistryMediumMCQMHT CET · 2023
What is the change in oxidation number of $Cr$ in the following redox reaction?
$3H_2O_2(aq) + Cr_2O_7^{2-}(aq) + 8H^+(aq) \rightarrow 3O_2(g) + 2Cr^{3+}(aq) + 7H_2O(l)$
$3H_2O_2(aq) + Cr_2O_7^{2-}(aq) + 8H^+(aq) \rightarrow 3O_2(g) + 2Cr^{3+}(aq) + 7H_2O(l)$
A
$+2$ to $+3$
B
$-2$ to $+3$
C
$+7$ to $+3$
D
$+6$ to $+3$
Solution
(D) In the given reaction,we need to find the change in the oxidation number of $Cr$ from the reactant side to the product side.
In the dichromate ion,$Cr_2O_7^{2-}$,let the oxidation number of $Cr$ be $x$.
$2x + 7(-2) = -2$
$2x - 14 = -2$
$2x = 12$
$x = +6$
In the product,the chromium ion is $Cr^{3+}$,so the oxidation number of $Cr$ is $+3$.
Therefore,the oxidation number of $Cr$ changes from $+6$ to $+3$.
In the dichromate ion,$Cr_2O_7^{2-}$,let the oxidation number of $Cr$ be $x$.
$2x + 7(-2) = -2$
$2x - 14 = -2$
$2x = 12$
$x = +6$
In the product,the chromium ion is $Cr^{3+}$,so the oxidation number of $Cr$ is $+3$.
Therefore,the oxidation number of $Cr$ changes from $+6$ to $+3$.
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162
ChemistryMediumMCQMHT CET · 2023
What is the change in oxidation number of selenium in the following redox reaction?
$SeO_{3(aq)}^{2-} + Cl_{2(g)} + 2OH^- \longrightarrow SeO_{4(aq)}^{2-} + 2Cl_{(aq)}^- + H_2O$
$SeO_{3(aq)}^{2-} + Cl_{2(g)} + 2OH^- \longrightarrow SeO_{4(aq)}^{2-} + 2Cl_{(aq)}^- + H_2O$
A
$+2$ to $-2$
B
$-2$ to $+2$
C
$+4$ to $+6$
D
$+3$ to $+4$
Solution
(C) To find the change in oxidation number of selenium $(Se)$,we calculate the oxidation state of $Se$ in both the reactant and product species.
The oxidation number of selenium changes from $+4$ to $+6$.
| Reactant: $SeO_3^{2-}$ | Product: $SeO_4^{2-}$ |
| Let the oxidation state of $Se$ be $x$. | Let the oxidation state of $Se$ be $x$. |
| $x + 3(-2) = -2$ | $x + 4(-2) = -2$ |
| $x - 6 = -2$ | $x - 8 = -2$ |
| $x = +4$ | $x = +6$ |
The oxidation number of selenium changes from $+4$ to $+6$.
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163
ChemistryMediumMCQMHT CET · 2023
What is the value of $x$ in order to balance the following redox reaction?
$Mn_{(aq)}^{2+} + x ClO_{3_{(aq)}}^{-} \rightarrow MnO_{2_{(s)}} + x ClO_{2_{(aq)}}$
$Mn_{(aq)}^{2+} + x ClO_{3_{(aq)}}^{-} \rightarrow MnO_{2_{(s)}} + x ClO_{2_{(aq)}}$
A
$x=1$
B
$x=2$
C
$x=3$
D
$x=4$
Solution
(B) To balance the redox reaction,we look at the change in oxidation states:
$Mn^{2+} \rightarrow MnO_2$: $Mn$ changes from $+2$ to $+4$ (oxidation,loss of $2e^-$).
$ClO_3^- \rightarrow ClO_2$: $Cl$ changes from $+5$ to $+4$ (reduction,gain of $1e^-$).
To balance the electrons,we multiply the reduction half-reaction by $2$:
$Mn^{2+} + 2 ClO_3^- \rightarrow MnO_2 + 2 ClO_2$.
Comparing this to the given equation $Mn^{2+} + x ClO_3^- \rightarrow MnO_2 + x ClO_2$,we find $x=2$.
$Mn^{2+} \rightarrow MnO_2$: $Mn$ changes from $+2$ to $+4$ (oxidation,loss of $2e^-$).
$ClO_3^- \rightarrow ClO_2$: $Cl$ changes from $+5$ to $+4$ (reduction,gain of $1e^-$).
To balance the electrons,we multiply the reduction half-reaction by $2$:
$Mn^{2+} + 2 ClO_3^- \rightarrow MnO_2 + 2 ClO_2$.
Comparing this to the given equation $Mn^{2+} + x ClO_3^- \rightarrow MnO_2 + x ClO_2$,we find $x=2$.
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164
ChemistryMediumMCQMHT CET · 2023
Calculate the amount of methane formed by the liberation of $149.6 \ kJ$ of heat using the following equation:
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ/mol$ (in $g$)
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ/mol$ (in $g$)
A
$16$
B
$24$
C
$32$
D
$48$
Solution
(C) According to the given thermochemical equation,$74.8 \ kJ$ of heat is evolved when $1 \ mol$ of methane $(CH_4)$ is formed.
The molar mass of $CH_4$ is $12 + (4 \times 1) = 16 \ g/mol$.
Thus,$74.8 \ kJ$ of heat corresponds to the formation of $16 \ g$ of $CH_4$.
For $149.6 \ kJ$ of heat,the amount of $CH_4$ formed is:
$x = \frac{149.6 \ kJ \times 16 \ g}{74.8 \ kJ} = 32 \ g$.
The molar mass of $CH_4$ is $12 + (4 \times 1) = 16 \ g/mol$.
Thus,$74.8 \ kJ$ of heat corresponds to the formation of $16 \ g$ of $CH_4$.
For $149.6 \ kJ$ of heat,the amount of $CH_4$ formed is:
$x = \frac{149.6 \ kJ \times 16 \ g}{74.8 \ kJ} = 32 \ g$.
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165
ChemistryEasyMCQMHT CET · 2023
What volume of $CO_{2(g)}$ at $STP$ is obtained by complete combustion of $6 \ g$ carbon (in $dm^3$)?
A
$22.4$
B
$11.2$
C
$5.6$
D
$2.24$
Solution
(B) The combustion reaction of carbon is: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
From the stoichiometry,$1 \ mol$ of $C$ produces $1 \ mol$ of $CO_{2(g)}$.
Molar mass of $C = 12 \ g/mol$.
Number of moles of $C = \frac{6 \ g}{12 \ g/mol} = 0.5 \ mol$.
Since $1 \ mol$ of $C$ gives $1 \ mol$ of $CO_{2(g)}$,$0.5 \ mol$ of $C$ will produce $0.5 \ mol$ of $CO_{2(g)}$.
At $STP$,the molar volume of an ideal gas is $22.4 \ dm^3/mol$.
Volume of $CO_{2(g)} = 0.5 \ mol \times 22.4 \ dm^3/mol = 11.2 \ dm^3$.
From the stoichiometry,$1 \ mol$ of $C$ produces $1 \ mol$ of $CO_{2(g)}$.
Molar mass of $C = 12 \ g/mol$.
Number of moles of $C = \frac{6 \ g}{12 \ g/mol} = 0.5 \ mol$.
Since $1 \ mol$ of $C$ gives $1 \ mol$ of $CO_{2(g)}$,$0.5 \ mol$ of $C$ will produce $0.5 \ mol$ of $CO_{2(g)}$.
At $STP$,the molar volume of an ideal gas is $22.4 \ dm^3/mol$.
Volume of $CO_{2(g)} = 0.5 \ mol \times 22.4 \ dm^3/mol = 11.2 \ dm^3$.
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166
ChemistryMediumMCQMHT CET · 2023
What is the mass of $KClO_{3(s)}$ required to liberate $22.4 \ dm^3$ of oxygen at $STP$ during thermal decomposition (in $g$)? (Molar Mass of $KClO_{3(s)} = 122.5 \ g/mol$)
A
$122.5$
B
$81.67$
C
$10.25$
D
$8.16$
Solution
(B) The balanced chemical equation for the thermal decomposition of potassium chlorate is: $2 KClO_{3(s)} \longrightarrow 2 KCl_{(s)} + 3 O_{2(g)}$
From the stoichiometry,$2 \text{ moles}$ of $KClO_3$ produce $3 \text{ moles}$ of $O_2$.
$2 \text{ moles}$ of $KClO_3 = 2 \times 122.5 \ g = 245 \ g$.
$3 \text{ moles}$ of $O_2$ at $STP$ occupy $3 \times 22.4 \ dm^3 = 67.2 \ dm^3$.
Thus,$245 \ g$ of $KClO_3$ produces $67.2 \ dm^3$ of $O_2$.
To find the mass $(x)$ required to produce $22.4 \ dm^3$ of $O_2$:
$x = \frac{245 \ g \times 22.4 \ dm^3}{67.2 \ dm^3} = 81.67 \ g$.
From the stoichiometry,$2 \text{ moles}$ of $KClO_3$ produce $3 \text{ moles}$ of $O_2$.
$2 \text{ moles}$ of $KClO_3 = 2 \times 122.5 \ g = 245 \ g$.
$3 \text{ moles}$ of $O_2$ at $STP$ occupy $3 \times 22.4 \ dm^3 = 67.2 \ dm^3$.
Thus,$245 \ g$ of $KClO_3$ produces $67.2 \ dm^3$ of $O_2$.
To find the mass $(x)$ required to produce $22.4 \ dm^3$ of $O_2$:
$x = \frac{245 \ g \times 22.4 \ dm^3}{67.2 \ dm^3} = 81.67 \ g$.
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167
ChemistryMediumMCQMHT CET · 2023
According to the reaction,$Mg_{(s)} + 2HCl_{(aq)} \longrightarrow MgCl_{2_{(aq)}} + H_{2_{(g)}} \uparrow$,calculate the mass of $Mg$ required to liberate $4.48 \ dm^3$ of $H_2$ gas at $STP$. (Molar mass of $Mg = 24 \ g \ mol^{-1}$) (in $g$)
A
$12$
B
$4.8$
C
$6$
D
$2.4$
Solution
(B) The number of moles of a gas $(n)$ is given by the formula: $n = \frac{\text{Volume of gas at STP}}{22.4 \ dm^3 \ mol^{-1}}$.
From the given volume,$n(H_2) = \frac{4.48 \ dm^3}{22.4 \ dm^3 \ mol^{-1}} = 0.2 \ mol$.
According to the balanced chemical equation: $Mg_{(s)} + 2HCl_{(aq)} \longrightarrow MgCl_{2_{(aq)}} + H_{2_{(g)}} \uparrow$.
$1 \ mol$ of $Mg$ produces $1 \ mol$ of $H_2$ gas.
Therefore,the moles of $Mg$ required to produce $0.2 \ mol$ of $H_2$ gas is $0.2 \ mol$.
Mass of $Mg = \text{moles} \times \text{molar mass} = 0.2 \ mol \times 24 \ g \ mol^{-1} = 4.8 \ g$.
From the given volume,$n(H_2) = \frac{4.48 \ dm^3}{22.4 \ dm^3 \ mol^{-1}} = 0.2 \ mol$.
According to the balanced chemical equation: $Mg_{(s)} + 2HCl_{(aq)} \longrightarrow MgCl_{2_{(aq)}} + H_{2_{(g)}} \uparrow$.
$1 \ mol$ of $Mg$ produces $1 \ mol$ of $H_2$ gas.
Therefore,the moles of $Mg$ required to produce $0.2 \ mol$ of $H_2$ gas is $0.2 \ mol$.
Mass of $Mg = \text{moles} \times \text{molar mass} = 0.2 \ mol \times 24 \ g \ mol^{-1} = 4.8 \ g$.
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168
ChemistryMediumMCQMHT CET · 2023
Which of the following statements is $NOT$ correct?
A
All alkali metals are silvery white.
B
Density of potassium is less than sodium.
C
Compounds of group-$1$ elements are diamagnetic.
D
Melting point of group-$1$ elements increases down the group.
Solution
(D) The melting point and boiling point of alkali metals (group-$1$ elements) decrease down the group because the metallic bond strength decreases as the atomic size increases. Therefore,the statement that the melting point increases down the group is incorrect.
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169
ChemistryMediumMCQMHT CET · 2023
Identify the $CORRECT$ decreasing order of melting point of cluster of sodium atoms depending on size.
A
Cluster of $10^3$ atoms $>$ cluster of $10^4$ atoms $>$ Bulk sodium
B
Bulk sodium $>$ cluster of $10^4$ atoms $>$ Cluster of $10^3$ atoms
C
Cluster of $10^4$ atoms $>$ Cluster of $10^3$ atoms $>$ Bulk sodium
D
Bulk sodium $>$ cluster of $10^3$ atoms $>$ Cluster of $10^4$ atoms
Solution
(B) The melting point of sodium clusters $(Na_n)$ increases as the number of atoms in the cluster increases.
Sodium clusters of $10^3$ atoms melt at $288 \ K$.
Sodium clusters of $10^4$ atoms melt at $303 \ K$.
Bulk sodium has a melting point of $371 \ K$.
Therefore,the decreasing order of melting point is: $\text{Bulk sodium} > \text{cluster of } 10^4 \text{ atoms} > \text{cluster of } 10^3 \text{ atoms}$.
Sodium clusters of $10^3$ atoms melt at $288 \ K$.
Sodium clusters of $10^4$ atoms melt at $303 \ K$.
Bulk sodium has a melting point of $371 \ K$.
Therefore,the decreasing order of melting point is: $\text{Bulk sodium} > \text{cluster of } 10^4 \text{ atoms} > \text{cluster of } 10^3 \text{ atoms}$.
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170
ChemistryEasyMCQMHT CET · 2023
Which among the following statements of group-$1$ elements is $NOT$ true?
A
Unipositive ions have inert gas configuration.
B
Compounds of unipositive ions are paramagnetic.
C
These form colourless compounds in $+1$ state.
D
These have high negative values of standard reduction potential.
Solution
(B) Group-$1$ elements (alkali metals) have a general electronic configuration of $ns^1$.
Upon losing one electron,they form unipositive ions $(M^+)$ with a stable noble gas configuration $(ns^2np^6)$,which are diamagnetic because all electrons are paired.
Therefore,the statement that compounds of unipositive ions are paramagnetic is incorrect.
They form colourless compounds in the $+1$ state because they lack unpaired electrons.
They are strong reducing agents and possess high negative values of standard reduction potential.
Upon losing one electron,they form unipositive ions $(M^+)$ with a stable noble gas configuration $(ns^2np^6)$,which are diamagnetic because all electrons are paired.
Therefore,the statement that compounds of unipositive ions are paramagnetic is incorrect.
They form colourless compounds in the $+1$ state because they lack unpaired electrons.
They are strong reducing agents and possess high negative values of standard reduction potential.
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171
ChemistryEasyMCQMHT CET · 2023
Which of the following is a property of alkali metals?
A
High density
B
Compounds are paramagnetic
C
Form dipositive ions only
D
Most electropositive elements
Solution
(D) Alkali metals are the most electropositive elements in the periodic table due to their low ionization enthalpy.
They have low density and form unipositive ions $(M^+)$ by losing their single valence electron.
Since they lose their only valence electron,the resulting ions have a noble gas configuration with no unpaired electrons,making their compounds diamagnetic.
They have low density and form unipositive ions $(M^+)$ by losing their single valence electron.
Since they lose their only valence electron,the resulting ions have a noble gas configuration with no unpaired electrons,making their compounds diamagnetic.
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172
ChemistryEasyMCQMHT CET · 2023
Identify the element having general electronic configuration $ns^1$ from the following.
A
$Ca$
B
$Sr$
C
$Ba$
D
$Fr$
Solution
(D) The general electronic configuration $ns^1$ corresponds to the group $1$ elements (alkali metals).
Among the given options,$Ca$,$Sr$,and $Ba$ belong to group $2$ (alkaline earth metals) with a general configuration of $ns^2$.
Francium $(Fr)$ belongs to group $1$,therefore it has the general electronic configuration $ns^1$.
Among the given options,$Ca$,$Sr$,and $Ba$ belong to group $2$ (alkaline earth metals) with a general configuration of $ns^2$.
Francium $(Fr)$ belongs to group $1$,therefore it has the general electronic configuration $ns^1$.
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173
ChemistryEasyMCQMHT CET · 2023
Which element from the following exhibits a common oxidation state of $+2$?
A
$Sr$
B
$Rb$
C
$Na$
D
$Li$
Solution
(A) The elements in Group $2$ of the periodic table,known as alkaline earth metals,typically exhibit a common oxidation state of $+2$.
Among the given options,$Sr$ (Strontium) belongs to Group $2$.
$Rb$ (Rubidium),$Na$ (Sodium),and $Li$ (Lithium) belong to Group $1$ (alkali metals) and typically exhibit an oxidation state of $+1$.
Among the given options,$Sr$ (Strontium) belongs to Group $2$.
$Rb$ (Rubidium),$Na$ (Sodium),and $Li$ (Lithium) belong to Group $1$ (alkali metals) and typically exhibit an oxidation state of $+1$.
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174
ChemistryEasyMCQMHT CET · 2023
Identify the alkaline earth metal from the following.
A
$Rb$
B
$Sr$
C
$Fr$
D
$Cs$
Solution
(B) The alkaline earth metals are the elements belonging to Group $2$ of the periodic table.
These elements have a general valence shell electronic configuration of $ns^2$.
Among the given options:
$Rb$ $(Rubidium)$ and $Cs$ $(Cesium)$ belong to Group $1$ (Alkali metals).
$Fr$ $(Francium)$ also belongs to Group $1$ (Alkali metals).
$Sr$ $(Strontium)$ belongs to Group $2$ (Alkaline earth metals).
Therefore,the correct option is $B$.
These elements have a general valence shell electronic configuration of $ns^2$.
Among the given options:
$Rb$ $(Rubidium)$ and $Cs$ $(Cesium)$ belong to Group $1$ (Alkali metals).
$Fr$ $(Francium)$ also belongs to Group $1$ (Alkali metals).
$Sr$ $(Strontium)$ belongs to Group $2$ (Alkaline earth metals).
Therefore,the correct option is $B$.
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175
ChemistryEasyMCQMHT CET · 2023
Which element from the following exhibits a diagonal relationship with beryllium $(Be)$?
A
$B$
B
$Na$
C
$Mg$
D
$Al$
Solution
(D) Beryllium $(Be)$ of the second period (Group $2$) shows a diagonal relationship with Aluminum $(Al)$ of the third period (Group $13$).
This is due to their similar ionic sizes and charge-to-radius ratios (ionic potential).
This is due to their similar ionic sizes and charge-to-radius ratios (ionic potential).
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176
ChemistryEasyMCQMHT CET · 2023
Which element from the following exhibits a diagonal relationship with $Mg$?
A
$Be$
B
$Li$
C
$Na$
D
$B$
Solution
(B) $Li$ and $Mg$ exhibit a diagonal relationship due to their similar ionic sizes and electronegativities.
This relationship occurs between certain elements of the second and third periods of the periodic table.
This relationship occurs between certain elements of the second and third periods of the periodic table.
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177
ChemistryEasyMCQMHT CET · 2023
Which of the following properties is exhibited by group $2$ elements?
A
Act as inert elements in $+1$ state.
B
Form $MH_2$ type hydrides with hydrogen on heating.
C
Elements at the top in the group catch fire when kept on water.
D
Reducing power of these elements is more than group $1$ elements.
Solution
(B) All the metals of group $2$,except beryllium,when heated with hydrogen form $MH_2$ type hydrides. Hence,statement $(B)$ is correct.
Group $2$ elements exhibit an oxidation state of $+2$.
Elements at the top of group $2$ do not catch fire when kept on water.
The reducing power of group $2$ elements is less than that of group $1$ elements due to higher ionization enthalpy. Hence,statements $(A)$,$(C)$,and $(D)$ are incorrect.
Group $2$ elements exhibit an oxidation state of $+2$.
Elements at the top of group $2$ do not catch fire when kept on water.
The reducing power of group $2$ elements is less than that of group $1$ elements due to higher ionization enthalpy. Hence,statements $(A)$,$(C)$,and $(D)$ are incorrect.
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178
ChemistryMediumMCQMHT CET · 2023
What is the concentration of $[H_3O^{+}]$ ion in $mol \ L^{-1}$ of $0.001 \ M$ acetic acid $(\alpha = 0.134)$?
A
$1.34 \times 10^{-4}$
B
$1.54 \times 10^{-4}$
C
$1.80 \times 10^{-4}$
D
$1.70 \times 10^{-4}$
Solution
(A) The concentration of $[H_3O^{+}]$ ions in a weak acid solution is given by the formula: $[H_3O^{+}] = \alpha \times c$
Given:
Concentration $(c) = 0.001 \ M = 10^{-3} \ M$
Degree of dissociation $(\alpha) = 0.134$
Substituting the values:
$[H_3O^{+}] = 0.134 \times 0.001$
$[H_3O^{+}] = 1.34 \times 10^{-4} \ mol \ L^{-1}$
Given:
Concentration $(c) = 0.001 \ M = 10^{-3} \ M$
Degree of dissociation $(\alpha) = 0.134$
Substituting the values:
$[H_3O^{+}] = 0.134 \times 0.001$
$[H_3O^{+}] = 1.34 \times 10^{-4} \ mol \ L^{-1}$
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179
ChemistryMediumMCQMHT CET · 2023
Acetic acid dissociates to $1.20 \%$ in its $0.01 \ M$ solution. What is the value of its dissociation constant?
A
$2.20 \times 10^{-2}$
B
$1.60 \times 10^{-4}$
C
$1.44 \times 10^{-6}$
D
$2.40 \times 10^{-4}$
Solution
(C) Percent dissociation $= 1.20 \%$
Degree of dissociation $(\alpha) = \frac{1.20}{100} = 0.012$
For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 c$
Substituting the values: $K_a = (0.012)^2 \times 0.01$
$K_a = 0.000144 \times 0.01 = 1.44 \times 10^{-6}$
Degree of dissociation $(\alpha) = \frac{1.20}{100} = 0.012$
For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 c$
Substituting the values: $K_a = (0.012)^2 \times 0.01$
$K_a = 0.000144 \times 0.01 = 1.44 \times 10^{-6}$
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180
ChemistryMCQMHT CET · 2023
Which one of the following aqueous solutions will exhibit the highest boiling point?
A
$0.01 \ M \ KNO_3$
B
$0.015 \ M$ urea
C
$0.015 \ M$ glucose
D
$0.01 \ M \ Na_2SO_4$
Solution
(D) Elevation in boiling point is a colligative property that depends on the number of solute particles in the solution.
Greater the number of solute particles,higher the elevation in boiling point.
We calculate the effective concentration $(i \cdot M)$ for each:
$A: 0.01 \ M \ KNO_3 \implies i = 2, \text{effective concentration} = 0.01 \times 2 = 0.02 \ M$
$B: 0.015 \ M \text{ urea} \implies i = 1, \text{effective concentration} = 0.015 \times 1 = 0.015 \ M$
$C: 0.015 \ M \text{ glucose} \implies i = 1, \text{effective concentration} = 0.015 \times 1 = 0.015 \ M$
$D: 0.01 \ M \ Na_2SO_4 \implies i = 3, \text{effective concentration} = 0.01 \times 3 = 0.03 \ M$
Since $0.01 \ M \ Na_2SO_4$ has the highest effective concentration of particles,it will exhibit the highest boiling point.
Greater the number of solute particles,higher the elevation in boiling point.
We calculate the effective concentration $(i \cdot M)$ for each:
$A: 0.01 \ M \ KNO_3 \implies i = 2, \text{effective concentration} = 0.01 \times 2 = 0.02 \ M$
$B: 0.015 \ M \text{ urea} \implies i = 1, \text{effective concentration} = 0.015 \times 1 = 0.015 \ M$
$C: 0.015 \ M \text{ glucose} \implies i = 1, \text{effective concentration} = 0.015 \times 1 = 0.015 \ M$
$D: 0.01 \ M \ Na_2SO_4 \implies i = 3, \text{effective concentration} = 0.01 \times 3 = 0.03 \ M$
Since $0.01 \ M \ Na_2SO_4$ has the highest effective concentration of particles,it will exhibit the highest boiling point.
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181
ChemistryMediumMCQMHT CET · 2023
Weak acid $HX$ has dissociation constant $1 \times 10^{-5}$. Calculate the percent dissociation in its $0.1 \ M$ solution. (in $\%$)
A
$2.2$
B
$3.5$
C
$4.2$
D
$1.0$
Solution
(D) Given: $K_{a} = 1 \times 10^{-5}$,$c = 0.1 \ M$.
For a weak acid,the dissociation constant is given by $K_{a} = \alpha^2 c$.
Therefore,$\alpha = \sqrt{\frac{K_{a}}{c}} = \sqrt{\frac{1 \times 10^{-5}}{0.1}} = \sqrt{1 \times 10^{-4}} = 0.01$.
Percent dissociation $= \alpha \times 100$.
Percent dissociation $= 0.01 \times 100 = 1.0 \%$.
For a weak acid,the dissociation constant is given by $K_{a} = \alpha^2 c$.
Therefore,$\alpha = \sqrt{\frac{K_{a}}{c}} = \sqrt{\frac{1 \times 10^{-5}}{0.1}} = \sqrt{1 \times 10^{-4}} = 0.01$.
Percent dissociation $= \alpha \times 100$.
Percent dissociation $= 0.01 \times 100 = 1.0 \%$.
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182
ChemistryEasyMCQMHT CET · 2023
Which one of the following sets of compounds correctly illustrates the law of reciprocal proportions?
A
$P_2O_3, PH_3$ and $H_2O$
B
$P_2O_5, PH_3$ and $H_2O$
C
$N_2O_5, NH_3$ and $H_2O$
D
$NO_2, NH_3$ and $H_2O$
Solution
(A) The law of reciprocal proportions states that if two different elements combine separately with a fixed mass of a third element,the ratio of the masses in which they do so is either the same or a simple multiple of the ratio of the masses in which they combine with each other.
In the set $P_2O_3, PH_3$ and $H_2O$:
$1$. $P$ combines with $H$ to form $PH_3$ ($31 \ g$ of $P$ with $3 \ g$ of $H$).
$2$. $P$ combines with $O$ to form $P_2O_3$ ($62 \ g$ of $P$ with $48 \ g$ of $O$).
$3$. For a fixed mass of $P$ $(62 \ g)$,the mass of $H$ is $6 \ g$ and the mass of $O$ is $48 \ g$. The ratio is $6:48 = 1:8$.
$4$. In $H_2O$,$H$ and $O$ combine in the ratio $2:16 = 1:8$.
Since the ratios are the same,this set illustrates the law.
In the set $P_2O_3, PH_3$ and $H_2O$:
$1$. $P$ combines with $H$ to form $PH_3$ ($31 \ g$ of $P$ with $3 \ g$ of $H$).
$2$. $P$ combines with $O$ to form $P_2O_3$ ($62 \ g$ of $P$ with $48 \ g$ of $O$).
$3$. For a fixed mass of $P$ $(62 \ g)$,the mass of $H$ is $6 \ g$ and the mass of $O$ is $48 \ g$. The ratio is $6:48 = 1:8$.
$4$. In $H_2O$,$H$ and $O$ combine in the ratio $2:16 = 1:8$.
Since the ratios are the same,this set illustrates the law.
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183
ChemistryEasyMCQMHT CET · 2023
What volume of ammonia is formed when $10 \ dm^3$ dinitrogen reacts with $30 \ dm^3$ dihydrogen at same temperature and pressure (in $dm^3$)?
A
$30$
B
$20$
C
$15$
D
$10$
Solution
(B) The balanced chemical equation for the reaction is:
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$
According to Avogadro's law,at constant temperature and pressure,the volume ratio of gaseous reactants and products is equal to their stoichiometric mole ratio.
From the equation,$1 \ volume$ of $N_2$ reacts with $3 \ volumes$ of $H_2$ to produce $2 \ volumes$ of $NH_3$.
Given $10 \ dm^3$ of $N_2$ and $30 \ dm^3$ of $H_2$,the stoichiometric ratio is exactly $1:3$.
Therefore,the volume of $NH_3$ formed is $2 \times 10 \ dm^3 = 20 \ dm^3$.
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$
According to Avogadro's law,at constant temperature and pressure,the volume ratio of gaseous reactants and products is equal to their stoichiometric mole ratio.
From the equation,$1 \ volume$ of $N_2$ reacts with $3 \ volumes$ of $H_2$ to produce $2 \ volumes$ of $NH_3$.
Given $10 \ dm^3$ of $N_2$ and $30 \ dm^3$ of $H_2$,the stoichiometric ratio is exactly $1:3$.
Therefore,the volume of $NH_3$ formed is $2 \times 10 \ dm^3 = 20 \ dm^3$.
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184
ChemistryEasyMCQMHT CET · 2023
What is the number of atoms present in $2.24 \ dm^3$ of $NH_3$ $(g)$ at $STP$?
A
$6.022 \times 10^{22}$
B
$2.4088 \times 10^{23}$
C
$1.8066 \times 10^{22}$
D
$6.022 \times 10^{23}$
Solution
(B) At $STP$,$22.4 \ dm^3$ of any gas contains $1 \ mol$ of molecules.
$1 \ mol$ of $NH_3$ contains $6.022 \times 10^{23}$ molecules.
Each $NH_3$ molecule contains $4$ atoms $(1 \ N + 3 \ H)$.
Therefore,$1 \ mol$ of $NH_3$ contains $4 \times 6.022 \times 10^{23} = 2.4088 \times 10^{24}$ atoms.
Given volume $= 2.24 \ dm^3$,which is $0.1 \ mol$ of $NH_3$.
Number of atoms $= 0.1 \times 2.4088 \times 10^{24} = 2.4088 \times 10^{23}$ atoms.
$1 \ mol$ of $NH_3$ contains $6.022 \times 10^{23}$ molecules.
Each $NH_3$ molecule contains $4$ atoms $(1 \ N + 3 \ H)$.
Therefore,$1 \ mol$ of $NH_3$ contains $4 \times 6.022 \times 10^{23} = 2.4088 \times 10^{24}$ atoms.
Given volume $= 2.24 \ dm^3$,which is $0.1 \ mol$ of $NH_3$.
Number of atoms $= 0.1 \times 2.4088 \times 10^{24} = 2.4088 \times 10^{23}$ atoms.
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185
ChemistryEasyMCQMHT CET · 2023
Which of the following pairs of compounds demonstrates the law of multiple proportions?
A
$CH_4, CCl_4$
B
$BF_3, NH_3$
C
$CO, CO_2$
D
$NO_2, CO_2$
Solution
(C) The law of multiple proportions states that when two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
In the pair $CO$ and $CO_2$,carbon $(C)$ and oxygen $(O)$ combine to form two different compounds.
In $CO$,$12 \ g$ of carbon combines with $16 \ g$ of oxygen.
In $CO_2$,$12 \ g$ of carbon combines with $32 \ g$ of oxygen.
The ratio of the masses of oxygen that combine with a fixed mass of carbon $(12 \ g)$ is $16:32$,which simplifies to $1:2$,a simple whole number ratio.
Therefore,$CO$ and $CO_2$ demonstrate the law of multiple proportions.
In the pair $CO$ and $CO_2$,carbon $(C)$ and oxygen $(O)$ combine to form two different compounds.
In $CO$,$12 \ g$ of carbon combines with $16 \ g$ of oxygen.
In $CO_2$,$12 \ g$ of carbon combines with $32 \ g$ of oxygen.
The ratio of the masses of oxygen that combine with a fixed mass of carbon $(12 \ g)$ is $16:32$,which simplifies to $1:2$,a simple whole number ratio.
Therefore,$CO$ and $CO_2$ demonstrate the law of multiple proportions.
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186
ChemistryMediumMCQMHT CET · 2023
What is the mass in gram of $1$ atom of an element if its atomic mass is $10 \ u$?
A
$2.06056 \times 10^{-22} \ g$
B
$1.66056 \times 10^{-23} \ g$
C
$1.06056 \times 10^{-24} \ g$
D
$3.66056 \times 10^{-25} \ g$
Solution
(B) Atomic mass is the mass of one atom of the element in atomic mass units $(u)$.
Given,mass of $1$ atom $= 10 \ u$.
We know that $1 \ u = 1.66056 \times 10^{-24} \ g$.
Therefore,mass of $1$ atom in grams $= 10 \times 1.66056 \times 10^{-24} \ g = 1.66056 \times 10^{-23} \ g$.
Given,mass of $1$ atom $= 10 \ u$.
We know that $1 \ u = 1.66056 \times 10^{-24} \ g$.
Therefore,mass of $1$ atom in grams $= 10 \times 1.66056 \times 10^{-24} \ g = 1.66056 \times 10^{-23} \ g$.
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187
ChemistryMediumMCQMHT CET · 2023
What is the number of moles of water molecules required to prepare $n$ moles of methane from $n$ moles of methyl magnesium iodide?
A
$n$
B
$2n$
C
$0.5n$
D
$0.1n$
Solution
(A) The chemical reaction between methyl magnesium iodide and water is given by:
$CH_3MgI + H_2O \rightarrow CH_4 + MgI(OH)$
From the balanced chemical equation,$1$ mole of methyl magnesium iodide reacts with $1$ mole of water to produce $1$ mole of methane.
Therefore,to prepare $n$ moles of methane from $n$ moles of methyl magnesium iodide,$n$ moles of water molecules are required.
$CH_3MgI + H_2O \rightarrow CH_4 + MgI(OH)$
From the balanced chemical equation,$1$ mole of methyl magnesium iodide reacts with $1$ mole of water to produce $1$ mole of methane.
Therefore,to prepare $n$ moles of methane from $n$ moles of methyl magnesium iodide,$n$ moles of water molecules are required.
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188
ChemistryMediumMCQMHT CET · 2023
What is the number of moles of $sp^2$ hybrid carbon atoms in one mole of $hexa-1,4-diyne$?
A
$5$
B
$3$
C
$4$
D
Zero
Solution
(D) The structure of $hexa-1,4-diyne$ is $CH_3-C \equiv C-CH_2-C \equiv C-H$.
In this molecule,the carbon atoms involved in triple bonds are $sp$ hybridized,and the carbon atoms involved in single bonds are $sp^3$ hybridized.
There are no carbon atoms with $sp^2$ hybridization in this molecule.
Therefore,the number of moles of $sp^2$ hybrid carbon atoms in one mole of $hexa-1,4-diyne$ is zero.
In this molecule,the carbon atoms involved in triple bonds are $sp$ hybridized,and the carbon atoms involved in single bonds are $sp^3$ hybridized.
There are no carbon atoms with $sp^2$ hybridization in this molecule.
Therefore,the number of moles of $sp^2$ hybrid carbon atoms in one mole of $hexa-1,4-diyne$ is zero.
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189
ChemistryMediumMCQMHT CET · 2023
What is the number of moles of $sp^2$ hybridized carbon atoms present in $n$ moles of isopentane?
A
zero
B
one
C
two
D
three
Solution
(A) The structure of isopentane is $CH_3-CH(CH_3)-CH_2-CH_3$.
In isopentane,all carbon atoms are bonded to four other atoms via single bonds.
Therefore,all carbon atoms in isopentane are $sp^3$ hybridized.
There are no $sp^2$ hybridized carbon atoms in isopentane.
Thus,the number of moles of $sp^2$ hybridized carbon atoms in $n$ moles of isopentane is $0$.
In isopentane,all carbon atoms are bonded to four other atoms via single bonds.
Therefore,all carbon atoms in isopentane are $sp^3$ hybridized.
There are no $sp^2$ hybridized carbon atoms in isopentane.
Thus,the number of moles of $sp^2$ hybridized carbon atoms in $n$ moles of isopentane is $0$.
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190
ChemistryMediumMCQMHT CET · 2023
How many moles of nitrogen atoms are present in $8 \ g$ of ammonium nitrate (in $mol$)?
A
$0.1$
B
$0.2$
C
$0.4$
D
$0.8$
Solution
(B) The chemical formula of ammonium nitrate is $NH_4NO_3$.
The molar mass of $NH_4NO_3 = (14 + 4 \times 1 + 14 + 3 \times 16) \ g/mol = 80 \ g/mol$.
Number of moles of $NH_4NO_3 = \frac{8 \ g}{80 \ g/mol} = 0.1 \ mol$.
Since $1 \ mol$ of $NH_4NO_3$ contains $2 \ mol$ of nitrogen atoms,
$0.1 \ mol$ of $NH_4NO_3$ contains $0.1 \times 2 = 0.2 \ mol$ of nitrogen atoms.
The molar mass of $NH_4NO_3 = (14 + 4 \times 1 + 14 + 3 \times 16) \ g/mol = 80 \ g/mol$.
Number of moles of $NH_4NO_3 = \frac{8 \ g}{80 \ g/mol} = 0.1 \ mol$.
Since $1 \ mol$ of $NH_4NO_3$ contains $2 \ mol$ of nitrogen atoms,
$0.1 \ mol$ of $NH_4NO_3$ contains $0.1 \times 2 = 0.2 \ mol$ of nitrogen atoms.
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191
ChemistryMediumMCQMHT CET · 2023
Which of the following substances consists of a total of $1 \ mol$ of atoms? (Molar mass of $NH_3 = 17 \ g/mol, H_2O = 18 \ g/mol, N_2 = 28 \ g/mol, CO_2 = 44 \ g/mol$)
A
$4.25 \ g$ of $NH_3$
B
$1.8 \ g$ of $H_2O$
C
$2.8 \ g$ of $N_2$
D
$4.4 \ g$ of $CO_2$
Solution
(A) For $4.25 \ g$ of $NH_3$: Moles of $NH_3 = \frac{4.25 \ g}{17 \ g/mol} = 0.25 \ mol$. Total atoms $= 0.25 \ mol \times (1 + 3) \ atoms/molecule = 1 \ mol$ of atoms.
For $1.8 \ g$ of $H_2O$: Moles of $H_2O = \frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$. Total atoms $= 0.1 \ mol \times (2 + 1) = 0.3 \ mol$ of atoms.
For $2.8 \ g$ of $N_2$: Moles of $N_2 = \frac{2.8 \ g}{28 \ g/mol} = 0.1 \ mol$. Total atoms $= 0.1 \ mol \times 2 = 0.2 \ mol$ of atoms.
For $4.4 \ g$ of $CO_2$: Moles of $CO_2 = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$. Total atoms $= 0.1 \ mol \times (1 + 2) = 0.3 \ mol$ of atoms.
Thus,$4.25 \ g$ of $NH_3$ contains $1 \ mol$ of atoms.
For $1.8 \ g$ of $H_2O$: Moles of $H_2O = \frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$. Total atoms $= 0.1 \ mol \times (2 + 1) = 0.3 \ mol$ of atoms.
For $2.8 \ g$ of $N_2$: Moles of $N_2 = \frac{2.8 \ g}{28 \ g/mol} = 0.1 \ mol$. Total atoms $= 0.1 \ mol \times 2 = 0.2 \ mol$ of atoms.
For $4.4 \ g$ of $CO_2$: Moles of $CO_2 = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$. Total atoms $= 0.1 \ mol \times (1 + 2) = 0.3 \ mol$ of atoms.
Thus,$4.25 \ g$ of $NH_3$ contains $1 \ mol$ of atoms.
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192
ChemistryEasyMCQMHT CET · 2023
What is the number of moles of secondary carbon atoms in $n \ mol$ of isopentane?
A
$4 n$
B
$3 n$
C
$2 n$
D
$n$
Solution
(D) The structure of isopentane is $CH_3-CH(CH_3)-CH_2-CH_3$.
In this molecule,there is only one secondary $(2^{\circ})$ carbon atom (the $-CH_2-$ group).
Therefore,$1 \ mol$ of isopentane contains $1 \ mol$ of secondary carbon atoms.
Thus,$n \ mol$ of isopentane contains $n \ mol$ of secondary carbon atoms.
In this molecule,there is only one secondary $(2^{\circ})$ carbon atom (the $-CH_2-$ group).
Therefore,$1 \ mol$ of isopentane contains $1 \ mol$ of secondary carbon atoms.
Thus,$n \ mol$ of isopentane contains $n \ mol$ of secondary carbon atoms.
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193
ChemistryMediumMCQMHT CET · 2023
What is the value of percent atom economy when an organic compound of formula weight $75 \ u$ is obtained from reactants having sum formula weight $225 \ u$?
A
$13.5$
B
$33.3$
C
$40.4$
D
$70.5$
Solution
(B) $\% \text{ atom economy} = \frac{\text{Formula weight of the desired product}}{\text{Sum of formula weight of all the reactants}} \times 100$
$\% \text{ atom economy} = \frac{75}{225} \times 100 = 33.3\%$
$\% \text{ atom economy} = \frac{75}{225} \times 100 = 33.3\%$
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194
ChemistryMediumMCQMHT CET · 2023
At $0^{\circ} C$,a gas occupies $22.4 \ L$. What is the temperature in Kelvin required to reach a volume of $224 \ L$ (in $K$)?
A
$546$
B
$273$
C
$2730$
D
$5460$
Solution
(C) According to Charles' law,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (at constant $n$ and $P$).
Given: $V_1 = 22.4 \ L$,$T_1 = 0^{\circ} C = 273 \ K$,$V_2 = 224 \ L$.
Substituting the values into the formula:
$T_2 = \frac{V_2 \times T_1}{V_1} = \frac{224 \times 273}{22.4} = 10 \times 273 = 2730 \ K$.
Given: $V_1 = 22.4 \ L$,$T_1 = 0^{\circ} C = 273 \ K$,$V_2 = 224 \ L$.
Substituting the values into the formula:
$T_2 = \frac{V_2 \times T_1}{V_1} = \frac{224 \times 273}{22.4} = 10 \times 273 = 2730 \ K$.
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195
ChemistryMediumMCQMHT CET · 2023
What is the new temperature of a gas when its initial volume $3 \ dm^3$ at $300 \ K$ is doubled at constant pressure (in $K$)?
A
$450$
B
$600$
C
$750$
D
$900$
Solution
(B) According to Charles' law,at constant pressure,the volume of a fixed mass of gas is directly proportional to its absolute temperature: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 3 \ dm^3$,$T_1 = 300 \ K$,and $V_2 = 2 \times V_1 = 6 \ dm^3$.
Substituting the values into the equation: $\frac{3 \ dm^3}{300 \ K} = \frac{6 \ dm^3}{T_2}$.
Solving for $T_2$: $T_2 = \frac{6 \ dm^3 \times 300 \ K}{3 \ dm^3} = 600 \ K$.
Given: $V_1 = 3 \ dm^3$,$T_1 = 300 \ K$,and $V_2 = 2 \times V_1 = 6 \ dm^3$.
Substituting the values into the equation: $\frac{3 \ dm^3}{300 \ K} = \frac{6 \ dm^3}{T_2}$.
Solving for $T_2$: $T_2 = \frac{6 \ dm^3 \times 300 \ K}{3 \ dm^3} = 600 \ K$.
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196
ChemistryMediumMCQMHT CET · 2023
What is the volume in $dm^3$ occupied by $3 \ mol$ of ammonia gas at $STP$?
A
$2.24$
B
$22.4$
C
$56.0$
D
$67.2$
Solution
(D) The volume of $1 \ mol$ of any ideal gas at $STP$ is $22.4 \ dm^3$.
Given number of moles of ammonia gas $(n) = 3 \ mol$.
Volume of gas at $STP = n \times 22.4 \ dm^3 \ mol^{-1}$.
Volume $= 3 \ mol \times 22.4 \ dm^3 \ mol^{-1} = 67.2 \ dm^3$.
Given number of moles of ammonia gas $(n) = 3 \ mol$.
Volume of gas at $STP = n \times 22.4 \ dm^3 \ mol^{-1}$.
Volume $= 3 \ mol \times 22.4 \ dm^3 \ mol^{-1} = 67.2 \ dm^3$.
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197
ChemistryEasyMCQMHT CET · 2023
If $N_2$ gas is compressed at $2 \text{ atm}$ from $9.0 \text{ L}$ to $3.0 \text{ L}$ at $300 \text{ K}$, find the final pressure at the same temperature. (in $\text{ atm}$)
A
$1.66$
B
$3.32$
C
$6.0$
D
$9.0$
Solution
(C) According to Boyle's law, for a fixed amount of gas at constant temperature, $P_1 V_1 = P_2 V_2$.
Given: $P_1 = 2 \text{ atm}$, $V_1 = 9.0 \text{ L}$, $V_2 = 3.0 \text{ L}$.
Substituting the values: $2 \text{ atm} \times 9.0 \text{ L} = P_2 \times 3.0 \text{ L}$.
$P_2 = \frac{2 \times 9.0}{3.0} \text{ atm} = 6.0 \text{ atm}$.
Given: $P_1 = 2 \text{ atm}$, $V_1 = 9.0 \text{ L}$, $V_2 = 3.0 \text{ L}$.
Substituting the values: $2 \text{ atm} \times 9.0 \text{ L} = P_2 \times 3.0 \text{ L}$.
$P_2 = \frac{2 \times 9.0}{3.0} \text{ atm} = 6.0 \text{ atm}$.
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198
ChemistryMediumMCQMHT CET · 2023
$A$ hot air balloon has a volume of $2000 \text{ dm}^3$ at $99^\circ\text{C}$. What is the new volume if the air in the balloon cools to $80^\circ\text{C}$ (in $\text{ dm}^3$)?
A
$2428.9$
B
$2656.9$
C
$2814.9$
D
$1897.8$
Solution
(D) According to Charles's Law, at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature $(V \propto T)$.
Given:
Initial volume $(V_1)$ = $2000 \text{ dm}^3$
Initial temperature $(T_1)$ = $99^\circ\text{C} = 99 + 273 = 372 \text{ K}$
Final temperature $(T_2)$ = $80^\circ\text{C} = 80 + 273 = 353 \text{ K}$
Using the formula: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$
$V_2 = \frac{V_1 \times T_2}{T_1}$
$V_2 = \frac{2000 \times 353}{372}$
$V_2 \approx 1897.8 \text{ dm}^3$.
Given:
Initial volume $(V_1)$ = $2000 \text{ dm}^3$
Initial temperature $(T_1)$ = $99^\circ\text{C} = 99 + 273 = 372 \text{ K}$
Final temperature $(T_2)$ = $80^\circ\text{C} = 80 + 273 = 353 \text{ K}$
Using the formula: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$
$V_2 = \frac{V_1 \times T_2}{T_1}$
$V_2 = \frac{2000 \times 353}{372}$
$V_2 \approx 1897.8 \text{ dm}^3$.
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199
ChemistryMediumMCQMHT CET · 2023
The volume of a gas is $4 \ dm^3$ at $0^{\circ} C$. Calculate the new volume at constant pressure when the temperature is increased by $10^{\circ} C$. (in $dm^3$)
A
$2.07$
B
$3.21$
C
$4.14$
D
$6.54$
Solution
(C) According to Charles' law,at constant pressure and amount of gas,the volume is directly proportional to the absolute temperature: $V_1 / T_1 = V_2 / T_2$.
Initial temperature $T_1 = 0^{\circ} C = 273 \ K$.
Final temperature $T_2 = 0^{\circ} C + 10^{\circ} C = 10^{\circ} C = 283 \ K$.
Given $V_1 = 4 \ dm^3$.
Substituting the values: $4 \ dm^3 / 273 \ K = V_2 / 283 \ K$.
$V_2 = (4 \times 283) / 273 \approx 4.1465 \ dm^3$.
Rounding to two decimal places,the volume is $4.14 \ dm^3$.
Initial temperature $T_1 = 0^{\circ} C = 273 \ K$.
Final temperature $T_2 = 0^{\circ} C + 10^{\circ} C = 10^{\circ} C = 283 \ K$.
Given $V_1 = 4 \ dm^3$.
Substituting the values: $4 \ dm^3 / 273 \ K = V_2 / 283 \ K$.
$V_2 = (4 \times 283) / 273 \approx 4.1465 \ dm^3$.
Rounding to two decimal places,the volume is $4.14 \ dm^3$.
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200
ChemistryDifficultMCQMHT CET · 2023
Find the temperature in degree Celsius if volume and pressure of $2 \ mol$ ideal gas is $20 \ dm^3$ and $4.926 \ atm$ respectively. $(R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1})$
A
$273$
B
$327$
C
$600$
D
$453$
Solution
(B) According to the ideal gas equation,$PV = nRT$.
Substituting the given values: $P = 4.926 \ atm$,$V = 20 \ dm^3$,$n = 2 \ mol$,and $R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
$T = \frac{PV}{nR} = \frac{4.926 \times 20}{2 \times 0.0821} = \frac{98.52}{0.1642} = 600 \ K$.
To convert the temperature to degree Celsius: $T(^{\circ}C) = T(K) - 273 = 600 - 273 = 327^{\circ}C$.
Substituting the given values: $P = 4.926 \ atm$,$V = 20 \ dm^3$,$n = 2 \ mol$,and $R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
$T = \frac{PV}{nR} = \frac{4.926 \times 20}{2 \times 0.0821} = \frac{98.52}{0.1642} = 600 \ K$.
To convert the temperature to degree Celsius: $T(^{\circ}C) = T(K) - 273 = 600 - 273 = 327^{\circ}C$.
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201
ChemistryEasyMCQMHT CET · 2023
What are the different possible oxidation states exhibited by scandium $(Sc)$?
A
$+4$
B
$+5$
C
$+4, +5$
D
$+3$
Solution
(D) Scandium ($Sc$,atomic number $21$) has the electronic configuration $[Ar] 3d^1 4s^2$.
It loses three electrons (two from $4s$ and one from $3d$) to achieve the stable noble gas configuration of Argon $([Ar])$.
Therefore,the only common and stable oxidation state exhibited by scandium is $+3$.
It loses three electrons (two from $4s$ and one from $3d$) to achieve the stable noble gas configuration of Argon $([Ar])$.
Therefore,the only common and stable oxidation state exhibited by scandium is $+3$.
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202
ChemistryMediumMCQMHT CET · 2023
Which of the following statements regarding transition elements is $NOT$ $CORRECT$?
A
These exhibit properties between $s$ and $p$ block elements.
B
These form cations with incomplete $d$-subshell.
C
$5d$ series consists of all elements from lanthanum to mercury of the periodic table.
D
These are arranged in four different $d$ series.
Solution
(C) The $5d$ series consists of elements from $Lanthanum$ ($La$,$Z=57$) and $Hafnium$ ($Hf$,$Z=72$) to $Mercury$ ($Hg$,$Z=80$). The statement that it consists of all elements from $La$ to $Hg$ is incorrect because the $4f$ series (lanthanoids) intervenes between $La$ and $Hf$.
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203
ChemistryEasyMCQMHT CET · 2023
What is the numerical value of spin only magnetic moment of copper in $+2$ state?
A
$0$
B
$1.73$
C
$2.78$
D
$4.4$
Solution
(B) The atomic number of copper $(Cu)$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
In the $+2$ oxidation state, $Cu^{2+}$ loses two electrons, resulting in the configuration $[Ar] 3d^9$.
In the $3d^9$ configuration, there is $n = 1$ unpaired electron.
The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 1$, we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
In the $+2$ oxidation state, $Cu^{2+}$ loses two electrons, resulting in the configuration $[Ar] 3d^9$.
In the $3d^9$ configuration, there is $n = 1$ unpaired electron.
The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 1$, we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
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204
ChemistryEasyMCQMHT CET · 2023
Which of the following is a non-ferrous alloy?
A
$A.$ Nickel steel
B
$B.$ Chromium steel
C
$C.$ Stainless steel
D
$D.$ Brass
Solution
(D) Brass is an alloy of copper $(Cu)$ and zinc $(Zn)$.
It does not contain iron $(Fe)$ as a primary component,making it a non-ferrous alloy.
Nickel steel,chromium steel,and stainless steel are all ferrous alloys as they contain iron.
It does not contain iron $(Fe)$ as a primary component,making it a non-ferrous alloy.
Nickel steel,chromium steel,and stainless steel are all ferrous alloys as they contain iron.
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205
ChemistryEasyMCQMHT CET · 2023
Which of the following nanoparticle catalysts is used in photocatalysis?
A
$TiO_2$
B
$Pd$
C
$Pt$
D
$Au$
Solution
(A) $TiO_2$ (Titanium dioxide) is a well-known semiconductor material that acts as an efficient photocatalyst under ultraviolet light irradiation.
It is widely used in environmental applications such as the degradation of organic pollutants and water splitting.
It is widely used in environmental applications such as the degradation of organic pollutants and water splitting.
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206
ChemistryEasyMCQMHT CET · 2023
Which of the following elements exhibits ferromagnetic properties?
A
$Mn$
B
$Co$
C
$Zn$
D
$Sc$
Solution
(B) Ferromagnetism is a property shown by substances that are strongly attracted by a magnetic field.
Among the transition metals,$Fe$,$Co$,and $Ni$ are known to exhibit ferromagnetic properties at room temperature.
Among the transition metals,$Fe$,$Co$,and $Ni$ are known to exhibit ferromagnetic properties at room temperature.
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207
ChemistryEasyMCQMHT CET · 2023
Which of the following alloys is used to make statues?
A
Nichrome
B
Stainless steel
C
Bronze
D
Cupra-nickel
Solution
(C) Bronze is an alloy primarily composed of copper $(Cu)$ and tin $(Sn)$.
It is widely used in the casting of statues and sculptures due to its durability,corrosion resistance,and ability to capture fine details.
It is widely used in the casting of statues and sculptures due to its durability,corrosion resistance,and ability to capture fine details.
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208
ChemistryEasyMCQMHT CET · 2023
Which of the following catalysts is used in the decomposition of $KClO_3$?
A
Platinized asbestos
B
$Fe-Cr$ catalyst
C
$Ni$
D
$MnO_2$
Solution
(D) The decomposition of potassium chlorate $(KClO_3)$ is a standard laboratory method for the preparation of oxygen gas.
The reaction is represented as: $2KClO_3(s) \xrightarrow{MnO_2, \Delta} 2KCl(s) + 3O_2(g)$.
In this reaction,manganese dioxide $(MnO_2)$ acts as a catalyst to lower the activation energy,allowing the decomposition to occur at a lower temperature.
The reaction is represented as: $2KClO_3(s) \xrightarrow{MnO_2, \Delta} 2KCl(s) + 3O_2(g)$.
In this reaction,manganese dioxide $(MnO_2)$ acts as a catalyst to lower the activation energy,allowing the decomposition to occur at a lower temperature.
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209
ChemistryEasyMCQMHT CET · 2023
Which among the following cations produces a colourless aqueous solution in its respective oxidation state?
A
$Ti^{3+}$
B
$V^{3+}$
C
$Sc^{3+}$
D
$Cu^{2+}$
Solution
(C) The electronic configuration of $Sc^{3+}$ is $[Ar] \ 3d^0$.
Since the $3d$ orbital is completely empty,there are no unpaired electrons present.
Because $d-d$ transitions are not possible in the absence of electrons in the $d$-orbital,the $Sc^{3+}$ ion forms a colourless aqueous solution.
Since the $3d$ orbital is completely empty,there are no unpaired electrons present.
Because $d-d$ transitions are not possible in the absence of electrons in the $d$-orbital,the $Sc^{3+}$ ion forms a colourless aqueous solution.
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210
ChemistryEasyMCQMHT CET · 2023
Which among the following elements does $NOT$ exhibit ferromagnetic properties?
A
$Cr$
B
$Fe$
C
$Co$
D
$Ni$
Solution
(A) Ferromagnetic substances are those which are strongly attracted by a magnetic field.
Common examples of ferromagnetic elements are $Fe$,$Co$,and $Ni$.
$Cr$ (Chromium) is an antiferromagnetic substance,not a ferromagnetic one.
Therefore,the correct answer is $Cr$.
Common examples of ferromagnetic elements are $Fe$,$Co$,and $Ni$.
$Cr$ (Chromium) is an antiferromagnetic substance,not a ferromagnetic one.
Therefore,the correct answer is $Cr$.
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211
ChemistryEasyMCQMHT CET · 2023
Identify the lanthanoid element from the following.
A
$Eu$
B
$Cm$
C
$Am$
D
$Np$
Solution
(A) The lanthanoids are the elements with atomic numbers $57$ to $71$.
$Eu$ (Europium) has an atomic number of $63$,which falls within the lanthanoid series.
$Cm$ (Curium),$Am$ (Americium),and $Np$ (Neptunium) are all actinoid elements (atomic numbers $89$ to $103$).
$Eu$ (Europium) has an atomic number of $63$,which falls within the lanthanoid series.
$Cm$ (Curium),$Am$ (Americium),and $Np$ (Neptunium) are all actinoid elements (atomic numbers $89$ to $103$).
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212
ChemistryMediumMCQMHT CET · 2023
Calculate $\Delta G^{\circ}$ for the cell: $Sn_{(s)} | Sn^{2+}_{(1M)} || Ag^{+}_{(1M)} | Ag_{(s)}$ at $25^{\circ} C$ given that $E^{\circ}_{cell} = 0.90 \ V$. (in $kJ$)
A
$-173.7$
B
$-225.3$
C
$-100.2$
D
$-290.8$
Solution
(A) The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
Here,the cell reaction is: $Sn_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$.
The number of electrons transferred,$n = 2$.
The Faraday constant,$F = 96500 \ C \ mol^{-1}$.
Given $E^{\circ}_{cell} = 0.90 \ V$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \times 0.90 \ J$.
$\Delta G^{\circ} = -173700 \ J = -173.7 \ kJ$.
Here,the cell reaction is: $Sn_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$.
The number of electrons transferred,$n = 2$.
The Faraday constant,$F = 96500 \ C \ mol^{-1}$.
Given $E^{\circ}_{cell} = 0.90 \ V$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \times 0.90 \ J$.
$\Delta G^{\circ} = -173700 \ J = -173.7 \ kJ$.
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213
ChemistryMediumMCQMHT CET · 2023
$A$ reaction,$Ni_{(s)} + Cu^{2+}_{(aq)} \rightarrow Ni^{2+}_{(aq)} + Cu_{(s)}$ occurs in a cell. Calculate $E^0_{cell}$ if $E^0_{Cu} = 0.337 \ V$ and $E^0_{Ni} = -0.257 \ V$. (in $V$)
A
$0.594$
B
$-0.594$
C
$-0.08$
D
$0.08$
Solution
(A) The standard cell potential is calculated using the formula:
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$
In the given reaction,$Ni$ is oxidized to $Ni^{2+}$ (anode) and $Cu^{2+}$ is reduced to $Cu$ (cathode).
Therefore,$E^0_{cell} = E^0_{Cu^{2+}/Cu} - E^0_{Ni^{2+}/Ni}$
$E^0_{cell} = 0.337 \ V - (-0.257 \ V)$
$E^0_{cell} = 0.337 \ V + 0.257 \ V = 0.594 \ V$
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$
In the given reaction,$Ni$ is oxidized to $Ni^{2+}$ (anode) and $Cu^{2+}$ is reduced to $Cu$ (cathode).
Therefore,$E^0_{cell} = E^0_{Cu^{2+}/Cu} - E^0_{Ni^{2+}/Ni}$
$E^0_{cell} = 0.337 \ V - (-0.257 \ V)$
$E^0_{cell} = 0.337 \ V + 0.257 \ V = 0.594 \ V$
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214
ChemistryEasyMCQMHT CET · 2023
Calculate $E_{\text{cell}}^{\circ}$ for the reaction: $Mg_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Mg_{(aq)}^{2+} + 2 Ag_{(s)}$,given that $E_{Ag^{+}/Ag}^{\circ} = 0.8 \ V$ and $E_{Mg^{2+}/Mg}^{\circ} = -2.37 \ V$. (in $V$)
A
$-3.17$
B
$3.17$
C
$-1.57$
D
$1.57$
Solution
(B) In the given cell reaction,$Mg$ undergoes oxidation at the anode and $Ag^{+}$ undergoes reduction at the cathode.
$E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$
$E_{\text{cell}}^{\circ} = E_{Ag^{+}/Ag}^{\circ} - E_{Mg^{2+}/Mg}^{\circ}$
$E_{\text{cell}}^{\circ} = 0.8 \ V - (-2.37 \ V) = 3.17 \ V$
$E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$
$E_{\text{cell}}^{\circ} = E_{Ag^{+}/Ag}^{\circ} - E_{Mg^{2+}/Mg}^{\circ}$
$E_{\text{cell}}^{\circ} = 0.8 \ V - (-2.37 \ V) = 3.17 \ V$
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215
ChemistryMediumMCQMHT CET · 2023
Find the number of faradays of electricity required to produce $45 \ g$ of $Al$ from molten $Al_2O_3$. (in $F$)
A
$1$
B
$3$
C
$5$
D
$7$
Solution
(C) The reduction reaction is: $Al^{3+} + 3e^{-} \longrightarrow Al$
This equation shows that $3 \ moles$ of electrons are required to produce $1 \ mole$ of $Al$,which is equivalent to $27 \ g$ of $Al$.
Thus,$3 \ F$ of electricity is required to produce $27 \ g$ of $Al$.
Therefore,the number of faradays of electricity required to produce $45 \ g$ of $Al$ is calculated as:
$\text{Faradays} = \frac{3 \ F}{27 \ g} \times 45 \ g = 5 \ F$.
This equation shows that $3 \ moles$ of electrons are required to produce $1 \ mole$ of $Al$,which is equivalent to $27 \ g$ of $Al$.
Thus,$3 \ F$ of electricity is required to produce $27 \ g$ of $Al$.
Therefore,the number of faradays of electricity required to produce $45 \ g$ of $Al$ is calculated as:
$\text{Faradays} = \frac{3 \ F}{27 \ g} \times 45 \ g = 5 \ F$.
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216
ChemistryEasyMCQMHT CET · 2023
Calculate the $E_{cell}^o$ for $Zn_{(s)}|Zn_{(1M)}^{2+}| |Cd_{(1M)}^{2+}|Cd_{(s)}$ at $25^{\circ} C$ given that $E_{Zn^{2+}/Zn}^{\circ} = -0.763 \ V$ and $E_{Cd^{2+}/Cd}^{\circ} = -0.403 \ V$. (in $V$)
A
$0.36$
B
$1.17$
C
$-0.36$
D
$-1.17$
Solution
(A) The cell reaction is represented as $Zn_{(s)}|Zn^{2+}_{(1M)}||Cd^{2+}_{(1M)}|Cd_{(s)}$.
In this cell,$Zn$ acts as the anode (oxidation) and $Cd$ acts as the cathode (reduction).
The standard cell potential is calculated using the formula:
$E_{cell}^o = E_{cathode}^{\circ} - E_{anode}^{\circ}$
Substituting the given values:
$E_{cell}^o = E_{Cd^{2+}/Cd}^{\circ} - E_{Zn^{2+}/Zn}^{\circ}$
$E_{cell}^o = -0.403 \ V - (-0.763 \ V)$
$E_{cell}^o = -0.403 \ V + 0.763 \ V = 0.36 \ V$.
In this cell,$Zn$ acts as the anode (oxidation) and $Cd$ acts as the cathode (reduction).
The standard cell potential is calculated using the formula:
$E_{cell}^o = E_{cathode}^{\circ} - E_{anode}^{\circ}$
Substituting the given values:
$E_{cell}^o = E_{Cd^{2+}/Cd}^{\circ} - E_{Zn^{2+}/Zn}^{\circ}$
$E_{cell}^o = -0.403 \ V - (-0.763 \ V)$
$E_{cell}^o = -0.403 \ V + 0.763 \ V = 0.36 \ V$.
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217
ChemistryEasyMCQMHT CET · 2023
Which of the following is the strongest reducing agent?
A
$K$
B
$Al$
C
$Mg$
D
$Ag$
Solution
(A) The strength of a reducing agent is determined by its standard electrode potential $(E^{\circ})$.
Lower (more negative) $E^{\circ}$ values indicate a stronger tendency to undergo oxidation,making the species a stronger reducing agent.
The standard reduction potentials are:
$K^+/K = -2.93 \ V$
$Mg^{2+}/Mg = -2.37 \ V$
$Al^{3+}/Al = -1.66 \ V$
$Ag^+/Ag = +0.80 \ V$
Since $K$ has the most negative $E^{\circ}$ value,it is the strongest reducing agent among the given options.
Lower (more negative) $E^{\circ}$ values indicate a stronger tendency to undergo oxidation,making the species a stronger reducing agent.
The standard reduction potentials are:
$K^+/K = -2.93 \ V$
$Mg^{2+}/Mg = -2.37 \ V$
$Al^{3+}/Al = -1.66 \ V$
$Ag^+/Ag = +0.80 \ V$
Since $K$ has the most negative $E^{\circ}$ value,it is the strongest reducing agent among the given options.
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218
ChemistryMediumMCQMHT CET · 2023
Electrolytic cells containing $Zn$ and $Al$ salt solutions are connected in series. If $6.5 \ g$ of $Zn$ is deposited in one cell,calculate the mass of $Al$ deposited in the second cell (molar mass: $Zn=65 \ g \ mol^{-1}$,$Al=27 \ g \ mol^{-1}$) by passing a definite quantity of electricity. (in $g$)
A
$2.4$
B
$2.1$
C
$2.7$
D
$1.8$
Solution
(D) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through cells connected in series,the mass of substances deposited is proportional to their equivalent masses.
$W_{Zn} / E_{Zn} = W_{Al} / E_{Al}$
Equivalent mass $E = \text{Molar mass} / n$-factor.
For $Zn^{2+} + 2e^{-} \longrightarrow Zn$,$n$-factor $= 2$.
$E_{Zn} = 65 / 2 = 32.5 \ g \ mol^{-1}$.
For $Al^{3+} + 3e^{-} \longrightarrow Al$,$n$-factor $= 3$.
$E_{Al} = 27 / 3 = 9 \ g \ mol^{-1}$.
Substituting the values:
$6.5 / 32.5 = W_{Al} / 9$
$0.2 = W_{Al} / 9$
$W_{Al} = 0.2 \times 9 = 1.8 \ g$.
$W_{Zn} / E_{Zn} = W_{Al} / E_{Al}$
Equivalent mass $E = \text{Molar mass} / n$-factor.
For $Zn^{2+} + 2e^{-} \longrightarrow Zn$,$n$-factor $= 2$.
$E_{Zn} = 65 / 2 = 32.5 \ g \ mol^{-1}$.
For $Al^{3+} + 3e^{-} \longrightarrow Al$,$n$-factor $= 3$.
$E_{Al} = 27 / 3 = 9 \ g \ mol^{-1}$.
Substituting the values:
$6.5 / 32.5 = W_{Al} / 9$
$0.2 = W_{Al} / 9$
$W_{Al} = 0.2 \times 9 = 1.8 \ g$.
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219
ChemistryMediumMCQMHT CET · 2023
Calculate $\Delta G^{\circ}$ for the reaction $Mg_{(s)} + Sn_{(aq)}^{2+} \longrightarrow Mg_{(aq)}^{2+} + Sn_{(s)}$ if $E_{cell}^{\circ}$ is $2.23 \ V$. (in $kJ$)
A
$-430.4$
B
$215.2$
C
$645.6$
D
$-860.8$
Solution
(A) The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE_{cell}^{\circ}$
Here,$n$ is the number of moles of electrons transferred in the balanced redox reaction. For $Mg + Sn^{2+} \longrightarrow Mg^{2+} + Sn$,$n = 2$.
$F$ is the Faraday constant,approximately $96500 \ C \ mol^{-1}$.
$E_{cell}^{\circ} = 2.23 \ V$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \times 2.23 \ J \ mol^{-1}$.
$\Delta G^{\circ} = -430390 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$ by dividing by $1000$: $\Delta G^{\circ} = -430.39 \ kJ \ mol^{-1} \approx -430.4 \ kJ \ mol^{-1}$.
Here,$n$ is the number of moles of electrons transferred in the balanced redox reaction. For $Mg + Sn^{2+} \longrightarrow Mg^{2+} + Sn$,$n = 2$.
$F$ is the Faraday constant,approximately $96500 \ C \ mol^{-1}$.
$E_{cell}^{\circ} = 2.23 \ V$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \times 2.23 \ J \ mol^{-1}$.
$\Delta G^{\circ} = -430390 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$ by dividing by $1000$: $\Delta G^{\circ} = -430.39 \ kJ \ mol^{-1} \approx -430.4 \ kJ \ mol^{-1}$.
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220
ChemistryEasyMCQMHT CET · 2023
Calculate the $E_{cell}$ for $Zn_{(s)} | Zn^{2+}_{(0.1 \ M)} || Cr^{3+}_{(0.1 \ M)} | Cr_{(s)}$ at $25^{\circ} C$ if $E^{\circ}_{cell}$ is $0.02 \ V$. (in $V$)
A
$-0.05$
B
$0.03$
C
$-0.06$
D
$0.07$
Solution
(B) The cell reaction is: $3 Zn_{(s)} + 2 Cr^{3+}_{(0.1 \ M)} \longrightarrow 3 Zn^{2+}_{(0.1 \ M)} + 2 Cr_{(s)}$
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592 \ V}{n} \log_{10} \frac{[Zn^{2+}]^3}{[Cr^{3+}]^2}$
Here,$n = 6$ (total electrons transferred).
$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \log_{10} \frac{(0.1)^3}{(0.1)^2}$
$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \log_{10} (0.1)$
Since $\log_{10} (0.1) = -1$,we have:
$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \times (-1)$
$E_{cell} = 0.02 \ V + 0.00987 \ V \approx 0.03 \ V$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592 \ V}{n} \log_{10} \frac{[Zn^{2+}]^3}{[Cr^{3+}]^2}$
Here,$n = 6$ (total electrons transferred).
$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \log_{10} \frac{(0.1)^3}{(0.1)^2}$
$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \log_{10} (0.1)$
Since $\log_{10} (0.1) = -1$,we have:
$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \times (-1)$
$E_{cell} = 0.02 \ V + 0.00987 \ V \approx 0.03 \ V$.
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221
ChemistryMediumMCQMHT CET · 2023
Calculate the time required in seconds to deposit $6.35 \ g$ of copper from its salt solution by passing a $5 \ A$ current. [Molar mass of $Cu = 63.5 \ g \ mol^{-1}$]
A
$3600$
B
$3700$
C
$3860$
D
$4000$
Solution
(C) The reduction reaction for copper is: $Cu^{2+} + 2e^{-} \longrightarrow Cu_{(s)}$
From the stoichiometry,$1 \ mol$ of $Cu$ requires $2 \ mol$ of electrons.
The amount of substance deposited is given by Faraday's law: $W = \frac{I \times t \times M}{n \times F}$
Where $W = 6.35 \ g$,$I = 5 \ A$,$M = 63.5 \ g \ mol^{-1}$,$n = 2$,and $F = 96500 \ C \ mol^{-1}$.
Substituting the values: $6.35 = \frac{5 \times t \times 63.5}{2 \times 96500}$
$t = \frac{6.35 \times 2 \times 96500}{5 \times 63.5}$
$t = \frac{0.1 \times 2 \times 96500}{5} = 3860 \ s$.
From the stoichiometry,$1 \ mol$ of $Cu$ requires $2 \ mol$ of electrons.
The amount of substance deposited is given by Faraday's law: $W = \frac{I \times t \times M}{n \times F}$
Where $W = 6.35 \ g$,$I = 5 \ A$,$M = 63.5 \ g \ mol^{-1}$,$n = 2$,and $F = 96500 \ C \ mol^{-1}$.
Substituting the values: $6.35 = \frac{5 \times t \times 63.5}{2 \times 96500}$
$t = \frac{6.35 \times 2 \times 96500}{5 \times 63.5}$
$t = \frac{0.1 \times 2 \times 96500}{5} = 3860 \ s$.
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222
ChemistryEasyMCQMHT CET · 2023
Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)
A
$0.36$
B
$-0.36$
C
$-0.18$
D
$0.18$
Solution
(A) The relationship between standard cell potential and equilibrium constant is given by the formula: $E_{cell}^{\circ} = \frac{0.0592}{n} \log_{10} K$ at $298 \ K$.
The given reaction is $2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$.
The half-reactions are:
Reduction: $Cu_{(aq)}^{+} + e^{-} \longrightarrow Cu_{(s)}$
Oxidation: $Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + e^{-}$
Thus,the number of electrons transferred,$n = 1$.
Substituting the values: $E_{cell}^{\circ} = \frac{0.0592}{1} \log_{10} (1.2 \times 10^6)$.
$E_{cell}^{\circ} = 0.0592 (\log 1.2 + \log 10^6)$.
$E_{cell}^{\circ} = 0.0592 (0.079 + 6)$.
$E_{cell}^{\circ} = 0.0592 \times 6.079 \approx 0.36 \ V$.
The given reaction is $2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$.
The half-reactions are:
Reduction: $Cu_{(aq)}^{+} + e^{-} \longrightarrow Cu_{(s)}$
Oxidation: $Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + e^{-}$
Thus,the number of electrons transferred,$n = 1$.
Substituting the values: $E_{cell}^{\circ} = \frac{0.0592}{1} \log_{10} (1.2 \times 10^6)$.
$E_{cell}^{\circ} = 0.0592 (\log 1.2 + \log 10^6)$.
$E_{cell}^{\circ} = 0.0592 (0.079 + 6)$.
$E_{cell}^{\circ} = 0.0592 \times 6.079 \approx 0.36 \ V$.
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223
ChemistryEasyMCQMHT CET · 2023
Calculate $E_{cell}^{\circ}$ for the following cell: $Zn_{(s)} | Zn^{2+}_{(1 \ M)} || Pb^{2+}_{(1 \ M)} | Pb_{(s)}$ given that $E^{\circ}_{Zn^{2+}/Zn} = -0.763 \ V$ and $E^{\circ}_{Pb^{2+}/Pb} = -0.126 \ V$. (in $V$)
A
$0.637$
B
$-0.530$
C
$-0.889$
D
$0.789$
Solution
(A) For the given cell,$Zn$ acts as the anode and $Pb$ acts as the cathode.
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$
$E_{cell}^{\circ} = E^{\circ}_{Pb^{2+}/Pb} - E^{\circ}_{Zn^{2+}/Zn}$
$E_{cell}^{\circ} = -0.126 \ V - (-0.763 \ V)$
$E_{cell}^{\circ} = -0.126 \ V + 0.763 \ V = 0.637 \ V$
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$
$E_{cell}^{\circ} = E^{\circ}_{Pb^{2+}/Pb} - E^{\circ}_{Zn^{2+}/Zn}$
$E_{cell}^{\circ} = -0.126 \ V - (-0.763 \ V)$
$E_{cell}^{\circ} = -0.126 \ V + 0.763 \ V = 0.637 \ V$
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224
ChemistryEasyMCQMHT CET · 2023
Which of the following expressions is used to find the cell potential of the $Cd_{(s)} | Cd_{(aq)}^{2+} || Cu_{(aq)}^{2+} | Cu_{(s)}$ cell at $25^{\circ} C$?
A
$E_{cell} = E_{cell}^{o} - 0.0296 \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$
B
$E_{cell} = E_{cell}^{o} + 0.0296 \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$
C
$E_{cell} = E_{cell}^{o} - 0.0592 \log \frac{[Cu^{2+}]}{[Cd^{2+}]}$
D
$E_{cell} = E_{cell}^{o} + 0.0592 \log \frac{[Cu^{2+}]}{[Cd^{2+}]}$
Solution
(A) The cell reaction is as follows:
Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$:
$E_{cell} = E_{cell}^{o} - \frac{0.0592}{n} \log \frac{[Product]}{[Reactant]}$
Here,the number of electrons transferred $n = 2$.
Substituting the concentrations:
$E_{cell} = E_{cell}^{o} - \frac{0.0592}{2} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$
$E_{cell} = E_{cell}^{o} - 0.0296 \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$
| $Cd_{(s)} \longrightarrow Cd_{(aq)}^{2+} + 2e^{-}$ | (Oxidation at anode) |
| $Cu_{(aq)}^{2+} + 2e^{-} \longrightarrow Cu_{(s)}$ | (Reduction at cathode) |
| $Cd_{(s)} + Cu_{(aq)}^{2+} \longrightarrow Cd_{(aq)}^{2+} + Cu_{(s)}$ | (Overall cell reaction) |
Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$:
$E_{cell} = E_{cell}^{o} - \frac{0.0592}{n} \log \frac{[Product]}{[Reactant]}$
Here,the number of electrons transferred $n = 2$.
Substituting the concentrations:
$E_{cell} = E_{cell}^{o} - \frac{0.0592}{2} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$
$E_{cell} = E_{cell}^{o} - 0.0296 \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$
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225
ChemistryEasyMCQMHT CET · 2023
Calculate $E_{\text{cell}}^{\circ}$ for $Cd_{(s)}|Cd^{2+}_{(1M)}||Ag^{+}_{(1M)}|Ag_{(s)}$. Given: $E^{\circ}_{Cd^{2+}/Cd} = -0.403 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.799 \ V$. (in $V$)
A
$1.202$
B
$-1.202$
C
$0.396$
D
$-0.396$
Solution
(A) For the given cell reaction,the anode is $Cd$ and the cathode is $Ag$.
The standard cell potential is calculated using the formula: $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$.
Substituting the given values: $E_{\text{cell}}^{\circ} = 0.799 \ V - (-0.403 \ V)$.
$E_{\text{cell}}^{\circ} = 0.799 + 0.403 = 1.202 \ V$.
The standard cell potential is calculated using the formula: $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$.
Substituting the given values: $E_{\text{cell}}^{\circ} = 0.799 \ V - (-0.403 \ V)$.
$E_{\text{cell}}^{\circ} = 0.799 + 0.403 = 1.202 \ V$.
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226
ChemistryEasyMCQMHT CET · 2023
Which among the following species is reduced by tin easily?
A
Iodine
B
Iron
C
Zinc
D
Sodium
Solution
(A) Among the given species,only iodine has a positive standard reduction potential $(E^{\circ})$ value.
Higher (more positive) $E^{\circ}$ value for a half-reaction indicates a greater tendency for the species to get reduced.
Since tin $(Sn)$ acts as a reducing agent,it will easily reduce the species with the highest reduction potential.
Therefore,iodine $(I_2)$ is reduced by tin $(Sn)$ easily.
Higher (more positive) $E^{\circ}$ value for a half-reaction indicates a greater tendency for the species to get reduced.
Since tin $(Sn)$ acts as a reducing agent,it will easily reduce the species with the highest reduction potential.
Therefore,iodine $(I_2)$ is reduced by tin $(Sn)$ easily.
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227
ChemistryEasyMCQMHT CET · 2023
Identify the gas produced due to reduction of $NH_4^{+}$ ions at cathode during working of dry cell.
A
Ammonia
B
Hydrogen
C
Hydrogen chloride
D
Chlorine
Solution
(B) In a dry cell,at the cathode,$NH_4^{+}$ ions undergo reduction to produce ammonia and hydrogen gas.
The reaction is: $2NH_{4(aq)}^{+} + 2e^{-} \longrightarrow 2NH_{3(aq)} + H_{2(g)}$.
Thus,the gas produced at the cathode is hydrogen.
The reaction is: $2NH_{4(aq)}^{+} + 2e^{-} \longrightarrow 2NH_{3(aq)} + H_{2(g)}$.
Thus,the gas produced at the cathode is hydrogen.
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228
ChemistryEasyMCQMHT CET · 2023
Identify the overall oxidation reaction that occurs in a lead storage cell during discharge.
A
$Pb_{(aq)}^{2+} + SO_{4(aq)}^{2-} \longrightarrow PbSO_{4(s)}$
B
$PbSO_{4(s)} + 2 H_2O_{(l)} \longrightarrow PbO_{2(s)} + 4 H_{(aq)}^{+} + SO_{4(aq)}^{2-} + 2 e^{-}$
C
$Pb_{(s)} + SO_{4(aq)}^{2-} \longrightarrow PbSO_{4(s)} + 2 e^{-}$
D
$PbSO_{4(s)} + 2 e^{-} \longrightarrow Pb_{(s)} + SO_{4(aq)}^{2-}$
Solution
(C) When the lead storage cell provides current (i.e.,during discharge),spongy lead $(Pb)$ is oxidized to $Pb^{2+}$ ions at the anode. The $Pb^{2+}$ ions so formed combine with $SO_4^{2-}$ ions from $H_2SO_4$ to form insoluble $PbSO_4$. The overall oxidation half-reaction is the sum of these two processes.
| $Pb_{(s)} \longrightarrow Pb_{(aq)}^{2+} + 2 e^{-}$ | (Oxidation) |
| $Pb_{(aq)}^{2+} + SO_{4(aq)}^{2-} \longrightarrow PbSO_{4(s)}$ | (Precipitation) |
| $Pb_{(s)} + SO_{4(aq)}^{2-} \longrightarrow PbSO_{4(s)} + 2 e^{-}$ | (Overall oxidation) |
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229
ChemistryEasyMCQMHT CET · 2023
Which of the following is $NOT$ true about a voltaic cell?
A
The anode acts as a negative electrode.
B
The cathode acts as a positive electrode.
C
It converts electrical energy into chemical energy.
D
$A$ dry cell is an example of a voltaic cell.
Solution
(C) voltaic cell (or galvanic cell) is an electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions.
Therefore,the statement that it converts electrical energy into chemical energy is incorrect.
In a voltaic cell,the anode is the negative electrode and the cathode is the positive electrode.
$A$ dry cell is a common type of voltaic cell.
Therefore,the statement that it converts electrical energy into chemical energy is incorrect.
In a voltaic cell,the anode is the negative electrode and the cathode is the positive electrode.
$A$ dry cell is a common type of voltaic cell.
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230
ChemistryMediumMCQMHT CET · 2023
Calculate the current in ampere required to deposit $4.8 \ g$ of $Cu$ from its salt solution in $30 \ minutes$. $\left[ \text{Molar mass of } Cu = 63.5 \ g \ mol^{-1} \right]$ (in $A$)
A
$8.1$
B
$6.4$
C
$10.5$
D
$12.3$
Solution
(A) The reduction reaction for copper is: $Cu^{2+} + 2e^{-} \longrightarrow Cu_{(s)}$
According to Faraday's law: $W = \frac{I \times t \times M}{n \times F}$
Where $W = 4.8 \ g$,$M = 63.5 \ g \ mol^{-1}$,$n = 2$,$t = 30 \times 60 \ s = 1800 \ s$,and $F = 96500 \ C \ mol^{-1}$.
Substituting the values: $4.8 = \frac{I \times 1800 \times 63.5}{2 \times 96500}$
$I = \frac{4.8 \times 2 \times 96500}{63.5 \times 1800}$
$I \approx 8.1 \ A$
According to Faraday's law: $W = \frac{I \times t \times M}{n \times F}$
Where $W = 4.8 \ g$,$M = 63.5 \ g \ mol^{-1}$,$n = 2$,$t = 30 \times 60 \ s = 1800 \ s$,and $F = 96500 \ C \ mol^{-1}$.
Substituting the values: $4.8 = \frac{I \times 1800 \times 63.5}{2 \times 96500}$
$I = \frac{4.8 \times 2 \times 96500}{63.5 \times 1800}$
$I \approx 8.1 \ A$
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231
ChemistryEasyMCQMHT CET · 2023
Which from the following compound solutions in water of equal concentration has electrical conductivity nearly same as distilled water?
A
Urea
B
Sodium chloride
C
Sodium hydroxide
D
Acetic acid
Solution
(A) Urea is a non-electrolyte and does not dissociate into ions in water.
Therefore,its electrical conductivity is nearly the same as that of distilled water.
Sodium chloride $(NaCl)$ and sodium hydroxide $(NaOH)$ are strong electrolytes,while acetic acid $(CH_3COOH)$ is a weak electrolyte,all of which increase the conductivity of water significantly.
Therefore,its electrical conductivity is nearly the same as that of distilled water.
Sodium chloride $(NaCl)$ and sodium hydroxide $(NaOH)$ are strong electrolytes,while acetic acid $(CH_3COOH)$ is a weak electrolyte,all of which increase the conductivity of water significantly.
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232
ChemistryMediumMCQMHT CET · 2023
What is the conductivity of $0.05 \ M \ BaCl_2$ solution if its molar conductivity is $220 \ \Omega^{-1} \ cm^2 \ mol^{-1}$?
A
$0.011 \ \Omega^{-1} \ cm^{-1}$
B
$0.022 \ \Omega^{-1} \ cm^{-1}$
C
$0.033 \ \Omega^{-1} \ cm^{-1}$
D
$0.044 \ \Omega^{-1} \ cm^{-1}$
Solution
(A) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(k)$ is given by the formula: $\Lambda_m = \frac{1000 \times k}{C}$
Rearranging for conductivity $(k)$: $k = \frac{\Lambda_m \times C}{1000}$
Given: $\Lambda_m = 220 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $C = 0.05 \ M$
Substituting the values: $k = \frac{220 \times 0.05}{1000} \ \Omega^{-1} \ cm^{-1}$
$k = \frac{11}{1000} \ \Omega^{-1} \ cm^{-1} = 0.011 \ \Omega^{-1} \ cm^{-1}$
Rearranging for conductivity $(k)$: $k = \frac{\Lambda_m \times C}{1000}$
Given: $\Lambda_m = 220 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $C = 0.05 \ M$
Substituting the values: $k = \frac{220 \times 0.05}{1000} \ \Omega^{-1} \ cm^{-1}$
$k = \frac{11}{1000} \ \Omega^{-1} \ cm^{-1} = 0.011 \ \Omega^{-1} \ cm^{-1}$
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233
ChemistryMediumMCQMHT CET · 2023
What mass of $Mg$ is produced during electrolysis of molten $MgCl_2$ by passing $2 \ A$ current for $482.5 \ s$ (in $g$)? $(Molar \ mass \ Mg = 24 \ g \ mol^{-1})$
A
$0.12$
B
$0.24$
C
$1.2$
D
$0.4$
Solution
(A) The reduction reaction at the cathode is: $Mg^{2+} + 2e^{-} \longrightarrow Mg_{(s)}$
According to Faraday's law,the mass $W$ deposited is given by: $W = \frac{I \times t \times M}{n \times F}$
Where $I = 2 \ A$,$t = 482.5 \ s$,$M = 24 \ g \ mol^{-1}$,$n = 2$,and $F = 96500 \ C \ mol^{-1}$.
$W = \frac{2 \times 482.5 \times 24}{2 \times 96500}$
$W = \frac{965 \times 24}{193000}$
$W = 0.12 \ g$
According to Faraday's law,the mass $W$ deposited is given by: $W = \frac{I \times t \times M}{n \times F}$
Where $I = 2 \ A$,$t = 482.5 \ s$,$M = 24 \ g \ mol^{-1}$,$n = 2$,and $F = 96500 \ C \ mol^{-1}$.
$W = \frac{2 \times 482.5 \times 24}{2 \times 96500}$
$W = \frac{965 \times 24}{193000}$
$W = 0.12 \ g$
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234
ChemistryMediumMCQMHT CET · 2023
Conductivity of a solution is $1.26 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$. Calculate molar conductivity for $0.01 \ M$ solution.
A
$1.26 \times 10^3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
B
$2.52 \times 10^3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
C
$4.82 \times 10^3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
D
$6.30 \times 10^3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Solution
(A) The formula for molar conductivity is $\Lambda_m = \frac{1000 \times \kappa}{c}$.
Given,conductivity $\kappa = 1.26 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$ and concentration $c = 0.01 \ M$.
Substituting the values:
$\Lambda_m = \frac{1000 \times 1.26 \times 10^{-2}}{0.01} \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$\Lambda_m = \frac{12.6}{0.01} \ \Omega^{-1} \ cm^2 \ mol^{-1} = 1.26 \times 10^3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Given,conductivity $\kappa = 1.26 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$ and concentration $c = 0.01 \ M$.
Substituting the values:
$\Lambda_m = \frac{1000 \times 1.26 \times 10^{-2}}{0.01} \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$\Lambda_m = \frac{12.6}{0.01} \ \Omega^{-1} \ cm^2 \ mol^{-1} = 1.26 \times 10^3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
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235
ChemistryMediumMCQMHT CET · 2023
Calculate the conductivity of $0.02 \ M$ electrolyte solution if its molar conductivity is $407.2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
A
$8.144 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
B
$4.072 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
C
$7.15 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
D
$6.055 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
Solution
(A) The relationship between molar conductivity $(\Lambda_{m})$ and conductivity $(k)$ is given by the formula: $\Lambda_{m} = \frac{1000 \times k}{c}$.
Rearranging the formula to solve for conductivity $(k)$: $k = \frac{\Lambda_{m} \times c}{1000}$.
Given values: $\Lambda_{m} = 407.2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $c = 0.02 \ M$.
Substituting the values: $k = \frac{407.2 \times 0.02}{1000}$.
$k = \frac{8.144}{1000} = 8.144 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.
Rearranging the formula to solve for conductivity $(k)$: $k = \frac{\Lambda_{m} \times c}{1000}$.
Given values: $\Lambda_{m} = 407.2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $c = 0.02 \ M$.
Substituting the values: $k = \frac{407.2 \times 0.02}{1000}$.
$k = \frac{8.144}{1000} = 8.144 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.
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236
ChemistryEasyMCQMHT CET · 2023
What happens when the solution of an electrolyte is diluted?
A
Both $\wedge$ and $k$ increase
B
Both $\wedge$ and $k$ decrease
C
$\wedge$ increases and $k$ decreases
D
$\wedge$ decreases and $k$ increases
Solution
(C) The molar conductivity $(\wedge_m)$ of an electrolyte increases upon dilution because the total volume of the solution containing one mole of electrolyte increases,leading to greater dissociation or increased ionic mobility.
Conversely,the conductivity $(k)$ (specific conductance) decreases upon dilution because the number of ions per unit volume of the solution decreases.
Conversely,the conductivity $(k)$ (specific conductance) decreases upon dilution because the number of ions per unit volume of the solution decreases.
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237
ChemistryEasyMCQMHT CET · 2023
Which among the following is the $CORRECT$ formula for the determination of cell constant?
A
$l/a = k/R$
B
$l/a = k \cdot R$
C
$l/a = R/k$
D
$l/a = 1/R$
Solution
(B) The conductivity $(k)$ is related to conductance $(G)$ and cell constant $(l/a)$ by the equation: $k = G \times (l/a)$.
Since conductance $G = 1/R$,we have $k = (1/R) \times (l/a)$.
Rearranging the formula to solve for the cell constant $(l/a)$: $l/a = k \times R$.
Since conductance $G = 1/R$,we have $k = (1/R) \times (l/a)$.
Rearranging the formula to solve for the cell constant $(l/a)$: $l/a = k \times R$.
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238
ChemistryMediumMCQMHT CET · 2023
The molar conductivity of $0.02 \ M$ $AgI$ at $298 \ K$ is $142.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. What is its conductivity?
A
$1.42 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
B
$2.41 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
C
$2.85 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
D
$7.11 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
Solution
(C) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(\kappa)$ is given by the formula:
$\Lambda_m = \frac{1000 \times \kappa}{c}$
Rearranging the formula to solve for conductivity $(\kappa)$:
$\kappa = \frac{\Lambda_m \times c}{1000}$
Given:
$\Lambda_m = 142.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$c = 0.02 \ M = 0.02 \ mol \ L^{-1}$
Substituting the values:
$\kappa = \frac{142.3 \times 0.02}{1000} \ \Omega^{-1} \ cm^{-1}$
$\kappa = 2.846 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
Rounding to two decimal places,we get:
$\kappa \approx 2.85 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
$\Lambda_m = \frac{1000 \times \kappa}{c}$
Rearranging the formula to solve for conductivity $(\kappa)$:
$\kappa = \frac{\Lambda_m \times c}{1000}$
Given:
$\Lambda_m = 142.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$c = 0.02 \ M = 0.02 \ mol \ L^{-1}$
Substituting the values:
$\kappa = \frac{142.3 \times 0.02}{1000} \ \Omega^{-1} \ cm^{-1}$
$\kappa = 2.846 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
Rounding to two decimal places,we get:
$\kappa \approx 2.85 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
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239
ChemistryEasyMCQMHT CET · 2023
$A$ conductivity cell containing $5 \times 10^{-4} \ M \ NaCl$ solution develops a resistance of $14000 \ \Omega$ at $25^{\circ} C$. Calculate the conductivity of the solution if the cell constant is $0.84 \ cm^{-1}$.
A
$6.0 \times 10^{-5} \ \Omega^{-1} \ cm^{-1}$
B
$3.0 \times 10^{-5} \ \Omega^{-1} \ cm^{-1}$
C
$9.0 \times 10^{-5} \ \Omega^{-1} \ cm^{-1}$
D
$12.0 \times 10^{-5} \ \Omega^{-1} \ cm^{-1}$
Solution
(A) The conductivity $(\kappa)$ is given by the formula: $\kappa = \frac{\text{cell constant}}{R}$.
Given,cell constant = $0.84 \ cm^{-1}$ and resistance $(R)$ = $14000 \ \Omega$.
Substituting the values: $\kappa = \frac{0.84 \ cm^{-1}}{14000 \ \Omega} = 6.0 \times 10^{-5} \ \Omega^{-1} \ cm^{-1}$.
Given,cell constant = $0.84 \ cm^{-1}$ and resistance $(R)$ = $14000 \ \Omega$.
Substituting the values: $\kappa = \frac{0.84 \ cm^{-1}}{14000 \ \Omega} = 6.0 \times 10^{-5} \ \Omega^{-1} \ cm^{-1}$.
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240
ChemistryEasyMCQMHT CET · 2023
Which of the following expressions represents the molar conductivity of an $AB_3$ type electrolyte?
A
$3 \lambda_{A^{3+}}^{\circ} + \lambda_{B^{-}}^{\circ}$
B
$\lambda_{A^{3+}}^{\circ} + \lambda_{B^{-}}^{\circ}$
C
$\lambda_{A^{3+}}^{\circ} + 3 \lambda_{B^{-}}^{\circ}$
D
$2 \lambda_{A^{3+}}^{\circ} + \lambda_{B^{-}}^{\circ}$
Solution
(C) The dissociation of an $AB_3$ type electrolyte is given by: $AB_3 \rightarrow A^{3+} + 3B^{-}$.
According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution $(\Lambda_m^{\circ})$ is the sum of the molar conductivities of the constituent ions multiplied by their respective stoichiometric coefficients.
Therefore,$\Lambda_m^{\circ} = \nu_{+} \lambda_{+}^{\circ} + \nu_{-} \lambda_{-}^{\circ}$.
Here,$\nu_{+} = 1$ (for $A^{3+}$) and $\nu_{-} = 3$ (for $B^{-}$).
Thus,$\Lambda_m^{\circ} = \lambda_{A^{3+}}^{\circ} + 3 \lambda_{B^{-}}^{\circ}$.
According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution $(\Lambda_m^{\circ})$ is the sum of the molar conductivities of the constituent ions multiplied by their respective stoichiometric coefficients.
Therefore,$\Lambda_m^{\circ} = \nu_{+} \lambda_{+}^{\circ} + \nu_{-} \lambda_{-}^{\circ}$.
Here,$\nu_{+} = 1$ (for $A^{3+}$) and $\nu_{-} = 3$ (for $B^{-}$).
Thus,$\Lambda_m^{\circ} = \lambda_{A^{3+}}^{\circ} + 3 \lambda_{B^{-}}^{\circ}$.
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241
ChemistryEasyMCQMHT CET · 2023
What is the molar conductivity of $0.005 \ M$ $NaI$ solution if its conductivity is $6.065 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$?
A
$121.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
B
$115.1 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
C
$126.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
D
$131.2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Solution
(A) The formula for molar conductivity is $\wedge_{m} = \frac{1000 \times \kappa}{c}$.
Given conductivity $\kappa = 6.065 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$ and concentration $c = 0.005 \ M$.
Substituting the values:
$\wedge_{m} = \frac{1000 \times 6.065 \times 10^{-4}}{0.005} = \frac{0.6065}{0.005} = 121.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Given conductivity $\kappa = 6.065 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$ and concentration $c = 0.005 \ M$.
Substituting the values:
$\wedge_{m} = \frac{1000 \times 6.065 \times 10^{-4}}{0.005} = \frac{0.6065}{0.005} = 121.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
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242
ChemistryMediumMCQMHT CET · 2023
Which of the following expressions represents the molar conductivity of an electrolyte of type $A_2B_3$?
A
$2 \lambda_{A^{3+}}^0 + 3 \lambda_{B^{2-}}^0$
B
$3 \lambda_{A^{2+}}^0 + 2 \lambda_{B^{3-}}^0$
C
$2 \lambda_{A^{3+}}^0 + 3 \lambda_{B^{2-}}^0$
D
$1 \lambda_{A^{6+}}^0 + 3 \lambda_{B^{2-}}^0$
Solution
(A) The molar conductivity at infinite dilution for an electrolyte is given by Kohlrausch's law: $\Lambda_m^0 = \nu_+ \lambda_+^0 + \nu_- \lambda_-^0$.
For the electrolyte $A_2B_3$,the dissociation reaction is: $A_2B_3 \rightarrow 2A^{3+} + 3B^{2-}$.
Here,the stoichiometric coefficients are $\nu_+ = 2$ and $\nu_- = 3$.
Therefore,the molar conductivity is: $\Lambda_m^0(A_2B_3) = 2 \lambda_{A^{3+}}^0 + 3 \lambda_{B^{2-}}^0$.
For the electrolyte $A_2B_3$,the dissociation reaction is: $A_2B_3 \rightarrow 2A^{3+} + 3B^{2-}$.
Here,the stoichiometric coefficients are $\nu_+ = 2$ and $\nu_- = 3$.
Therefore,the molar conductivity is: $\Lambda_m^0(A_2B_3) = 2 \lambda_{A^{3+}}^0 + 3 \lambda_{B^{2-}}^0$.
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243
ChemistryEasyMCQMHT CET · 2023
Calculate the molar conductivity of $NH_4OH$ at infinite dilution if the molar conductivities of $Ba(OH)_2$,$BaCl_2$,and $NH_4Cl$ at infinite dilution are $520$,$280$,and $129 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.
A
$249.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
B
$498.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
C
$125.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
D
$369.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Solution
(A) According to Kohlrausch's law of independent migration of ions:
$\wedge_m^0(NH_4OH) = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0$
We are given:
$1. \wedge_m^0(Ba(OH)_2) = \lambda_{Ba^{2+}}^0 + 2\lambda_{OH^-}^0 = 520 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$2. \wedge_m^0(BaCl_2) = \lambda_{Ba^{2+}}^0 + 2\lambda_{Cl^-}^0 = 280 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$3. \wedge_m^0(NH_4Cl) = \lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0 = 129 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
To get $\wedge_m^0(NH_4OH)$,we perform the operation: $\frac{1}{2} \wedge_m^0(Ba(OH)_2) + \wedge_m^0(NH_4Cl) - \frac{1}{2} \wedge_m^0(BaCl_2)$
$= \frac{1}{2}(520) + 129 - \frac{1}{2}(280)$
$= 260 + 129 - 140$
$= 249.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\wedge_m^0(NH_4OH) = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0$
We are given:
$1. \wedge_m^0(Ba(OH)_2) = \lambda_{Ba^{2+}}^0 + 2\lambda_{OH^-}^0 = 520 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$2. \wedge_m^0(BaCl_2) = \lambda_{Ba^{2+}}^0 + 2\lambda_{Cl^-}^0 = 280 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$3. \wedge_m^0(NH_4Cl) = \lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0 = 129 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
To get $\wedge_m^0(NH_4OH)$,we perform the operation: $\frac{1}{2} \wedge_m^0(Ba(OH)_2) + \wedge_m^0(NH_4Cl) - \frac{1}{2} \wedge_m^0(BaCl_2)$
$= \frac{1}{2}(520) + 129 - \frac{1}{2}(280)$
$= 260 + 129 - 140$
$= 249.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
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244
ChemistryEasyMCQMHT CET · 2023
$A$ conductivity cell containing $0.001 \ M$ $AgNO_3$ solution develops resistance $6530 \ \Omega$ at $25 \ ^{\circ}C$. Calculate the electrical conductivity of the solution at the same temperature if the cell constant is $0.653 \ cm^{-1}$.
A
$1.3 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$
B
$1.5 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$
C
$1.7 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$
D
$1.0 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$
Solution
(D) The relationship between conductivity $(k)$,cell constant $(G^*)$,and resistance $(R)$ is given by:
$k = \frac{G^*}{R}$
Given:
$G^* = 0.653 \ cm^{-1}$
$R = 6530 \ \Omega$
Substituting the values:
$k = \frac{0.653 \ cm^{-1}}{6530 \ \Omega} = 1 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$
$k = \frac{G^*}{R}$
Given:
$G^* = 0.653 \ cm^{-1}$
$R = 6530 \ \Omega$
Substituting the values:
$k = \frac{0.653 \ cm^{-1}}{6530 \ \Omega} = 1 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$
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245
ChemistryEasyMCQMHT CET · 2023
What is the common name of $Benzene-1,3-diol$?
A
Catechol
B
Resorcinol
C
Quinol
D
Pyrogallol
Solution
(B) The structure of $Benzene-1,3-diol$ consists of a benzene ring with two hydroxyl $(-OH)$ groups attached at the $1$ and $3$ positions.
This compound is commonly known as $Resorcinol$.
$Catechol$ is $Benzene-1,2-diol$.
$Quinol$ (or $Hydroquinone$) is $Benzene-1,4-diol$.
$Pyrogallol$ is $Benzene-1,2,3-triol$.
This compound is commonly known as $Resorcinol$.
$Catechol$ is $Benzene-1,2-diol$.
$Quinol$ (or $Hydroquinone$) is $Benzene-1,4-diol$.
$Pyrogallol$ is $Benzene-1,2,3-triol$.
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246
ChemistryEasyMCQMHT CET · 2023
What is the $IUPAC$ name of propylene glycol?
A
Propane-$1,2$-diol
B
Propane-$1,3$-diol
C
Propane-$1,2,3$-triol
D
Propene-$1,2,3$-triol
Solution
(A) The common name 'propylene glycol' refers to the chemical compound $CH_3-CH(OH)-CH_2OH$.
Its $IUPAC$ name is determined by identifying the longest carbon chain containing the hydroxyl groups,which is a propane chain ($3$ carbons).
The hydroxyl groups are attached at positions $1$ and $2$.
Therefore,the $IUPAC$ name is Propane-$1,2$-diol.
Its $IUPAC$ name is determined by identifying the longest carbon chain containing the hydroxyl groups,which is a propane chain ($3$ carbons).
The hydroxyl groups are attached at positions $1$ and $2$.
Therefore,the $IUPAC$ name is Propane-$1,2$-diol.
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247
ChemistryEasyMCQMHT CET · 2023
Which among the following phenols does $NOT$ correctly match with their $IUPAC$ names?
A
Catechol : Benzene$-1,2-$diol
B
Resorcinol : Benzene$-1,3-$diol
C
$o-$Cresol : Benzene$-1,2,3-$triol
D
Quinol : Benzene$-1,4-$diol
Solution
(C) The $IUPAC$ names for the given compounds are as follows:
$1$. Catechol is $Benzene-1,2-diol$.
$2$. Resorcinol is $Benzene-1,3-diol$.
$3$. $o-Cresol$ is $2-Methylphenol$.
$4$. Quinol is $Benzene-1,4-diol$.
Comparing these with the options,$o-Cresol$ is incorrectly matched with $Benzene-1,2,3-triol$. Therefore,option $C$ is the correct answer.
$1$. Catechol is $Benzene-1,2-diol$.
$2$. Resorcinol is $Benzene-1,3-diol$.
$3$. $o-Cresol$ is $2-Methylphenol$.
$4$. Quinol is $Benzene-1,4-diol$.
Comparing these with the options,$o-Cresol$ is incorrectly matched with $Benzene-1,2,3-triol$. Therefore,option $C$ is the correct answer.
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248
ChemistryEasyMCQMHT CET · 2023
What is the $IUPAC$ name of the compound shown in the image?
A
$1-$Bromo$-4-$hydroxybenzene
B
$1-$Hydroxy$-4-$bromobenzene
C
$4-$Hydroxybromobenzene
D
$4-$Bromophenol
Solution
(D) The given compound is a benzene ring substituted with a hydroxyl group $(-OH)$ and a bromine atom $(-Br)$ at the para position relative to each other.
According to $IUPAC$ nomenclature rules,the hydroxyl group attached to a benzene ring is given priority,and the parent compound is named as phenol.
The carbon atom attached to the $-OH$ group is assigned position $1$.
Counting from this position,the bromine atom is at position $4$.
Therefore,the $IUPAC$ name is $4-$bromophenol.
According to $IUPAC$ nomenclature rules,the hydroxyl group attached to a benzene ring is given priority,and the parent compound is named as phenol.
The carbon atom attached to the $-OH$ group is assigned position $1$.
Counting from this position,the bromine atom is at position $4$.
Therefore,the $IUPAC$ name is $4-$bromophenol.
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249
ChemistryMediumMCQMHT CET · 2023
Which of the following is the correct $IUPAC$ name of catechol?
A
Benzene-$1,2$-diol
B
Benzene-$1,3$-diol
C
Benzene-$1,4$-diol
D
Benzene-$1,3,5$-triol
Solution
(A) Catechol is a common name for a dihydroxybenzene derivative where the two hydroxyl $(-OH)$ groups are attached to adjacent carbon atoms on the benzene ring.
According to $IUPAC$ nomenclature rules,the benzene ring is numbered such that the substituents get the lowest possible locants.
For catechol,the $-OH$ groups are at positions $1$ and $2$.
Therefore,the correct $IUPAC$ name is Benzene-$1,2$-diol.
According to $IUPAC$ nomenclature rules,the benzene ring is numbered such that the substituents get the lowest possible locants.
For catechol,the $-OH$ groups are at positions $1$ and $2$.
Therefore,the correct $IUPAC$ name is Benzene-$1,2$-diol.
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250
ChemistryEasyMCQMHT CET · 2023
Which of the following is a secondary benzylic alcohol?
A
$C_6H_5CH_2OH$
B
$C_6H_5CH(OH)CH_3$
C
$C_6H_5C(CH_3)_2OH$
D
$C_6H_5C(CH_3)(C_2H_5)OH$
Solution
(B) benzylic alcohol is one where the $-OH$ group is attached to a carbon atom next to an aromatic ring.
$1$. In $C_6H_5CH_2OH$,the carbon attached to $-OH$ is bonded to one carbon (the ring) and two hydrogens,making it a primary $(1^{\circ})$ benzylic alcohol.
$2$. In $C_6H_5CH(OH)CH_3$,the carbon attached to $-OH$ is bonded to one carbon (the ring) and one methyl group,making it a secondary $(2^{\circ})$ benzylic alcohol.
$3$. In $C_6H_5C(CH_3)_2OH$,the carbon attached to $-OH$ is bonded to the ring and two methyl groups,making it a tertiary $(3^{\circ})$ benzylic alcohol.
$4$. In $C_6H_5C(CH_3)(C_2H_5)OH$,the carbon attached to $-OH$ is bonded to the ring,one methyl,and one ethyl group,making it a tertiary $(3^{\circ})$ benzylic alcohol.
Thus,the correct option is $B$.
$1$. In $C_6H_5CH_2OH$,the carbon attached to $-OH$ is bonded to one carbon (the ring) and two hydrogens,making it a primary $(1^{\circ})$ benzylic alcohol.
$2$. In $C_6H_5CH(OH)CH_3$,the carbon attached to $-OH$ is bonded to one carbon (the ring) and one methyl group,making it a secondary $(2^{\circ})$ benzylic alcohol.
$3$. In $C_6H_5C(CH_3)_2OH$,the carbon attached to $-OH$ is bonded to the ring and two methyl groups,making it a tertiary $(3^{\circ})$ benzylic alcohol.
$4$. In $C_6H_5C(CH_3)(C_2H_5)OH$,the carbon attached to $-OH$ is bonded to the ring,one methyl,and one ethyl group,making it a tertiary $(3^{\circ})$ benzylic alcohol.
Thus,the correct option is $B$.
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