ChemistryQ151–250 of 627 questions
Page 4 of 8 · English
151
ChemistryMediumMCQMHT CET · 2022
What is the percentage dissociation of $0.1 \ M$ acetic acid (in $\%$)? $(K_a = 10^{-5})$
A
$1$
B
$10$
C
$0.1$
D
$0.01$
Solution
(A) For a weak acid,the dissociation constant $K_a$ is related to the degree of dissociation $\alpha$ and concentration $C$ by the formula: $K_a = \alpha^2 C$.
Given $K_a = 10^{-5}$ and $C = 0.1 \ M$.
Substituting the values: $10^{-5} = \alpha^2 \times 0.1$.
$\alpha^2 = \frac{10^{-5}}{0.1} = 10^{-4}$.
$\alpha = \sqrt{10^{-4}} = 0.01$.
Percentage dissociation = $\alpha \times 100 = 0.01 \times 100 = 1 \%$.
Given $K_a = 10^{-5}$ and $C = 0.1 \ M$.
Substituting the values: $10^{-5} = \alpha^2 \times 0.1$.
$\alpha^2 = \frac{10^{-5}}{0.1} = 10^{-4}$.
$\alpha = \sqrt{10^{-4}} = 0.01$.
Percentage dissociation = $\alpha \times 100 = 0.01 \times 100 = 1 \%$.
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152
ChemistryMediumMCQMHT CET · 2022
What is the quantity of sugar charcoal obtained when $34.2 \ g$ of sugar is charred using the required quantity of concentrated sulphuric acid under ideal conditions (in $g$)?
A
$114$
B
$14.4$
C
$11.0$
D
$10.5$
Solution
(B) The dehydration reaction of sugar $(C_{12}H_{22}O_{11})$ by concentrated sulphuric acid is given by:
$C_{12}H_{22}O_{11} \rightarrow 12C + 11H_2O$
The molar mass of sugar $(C_{12}H_{22}O_{11})$ is $12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342 \ g/mol$.
Number of moles of sugar = $\frac{34.2 \ g}{342 \ g/mol} = 0.1 \ mol$.
From the stoichiometry of the reaction,$1 \ mol$ of sugar produces $12 \ mol$ of carbon (sugar charcoal).
Therefore,$0.1 \ mol$ of sugar produces $0.1 \times 12 = 1.2 \ mol$ of carbon.
Mass of charcoal = $1.2 \ mol \times 12 \ g/mol = 14.4 \ g$.
$C_{12}H_{22}O_{11} \rightarrow 12C + 11H_2O$
The molar mass of sugar $(C_{12}H_{22}O_{11})$ is $12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342 \ g/mol$.
Number of moles of sugar = $\frac{34.2 \ g}{342 \ g/mol} = 0.1 \ mol$.
From the stoichiometry of the reaction,$1 \ mol$ of sugar produces $12 \ mol$ of carbon (sugar charcoal).
Therefore,$0.1 \ mol$ of sugar produces $0.1 \times 12 = 1.2 \ mol$ of carbon.
Mass of charcoal = $1.2 \ mol \times 12 \ g/mol = 14.4 \ g$.
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153
ChemistryMediumMCQMHT CET · 2022
Mass of one molecule of oxygen in $amu$ and in $gram$ respectively is
A
$32 \ u, 53.13 \times 10^{-24} \ g$
B
$16 \ u, 6.0 \times 10^{-24} \ g$
C
$42 \ u, 5.313 \times 10^{-24} \ g$
D
$53.13 \times 10^{-24} \ u, 32 \ g$
Solution
(A) The molar mass of an oxygen molecule $(O_2)$ is $32 \ g/mol$.
The mass of one molecule in $amu$ is equal to its molecular mass,which is $32 \ u$.
To convert this to grams,we use the conversion factor $1 \ u = 1.66056 \times 10^{-24} \ g$.
Mass in grams $= 32 \times 1.66056 \times 10^{-24} \ g \approx 53.14 \times 10^{-24} \ g$.
The mass of one molecule in $amu$ is equal to its molecular mass,which is $32 \ u$.
To convert this to grams,we use the conversion factor $1 \ u = 1.66056 \times 10^{-24} \ g$.
Mass in grams $= 32 \times 1.66056 \times 10^{-24} \ g \approx 53.14 \times 10^{-24} \ g$.
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154
ChemistryMediumMCQMHT CET · 2022
How many moles of oxygen gas at $STP$ are equivalent to $5.6 \ L$?
A
$\frac{1}{8} \text{ mole}$
B
$1 \text{ mole}$
C
$\frac{1}{2} \text{ mole}$
D
$\frac{1}{4} \text{ mole}$
Solution
(D) At $STP$,$1 \text{ mole}$ of any gas occupies $22.4 \ L$ volume.
Number of moles $= \frac{\text{Given volume}}{\text{Molar volume at } STP} = \frac{5.6 \ L}{22.4 \ L/mol} = 0.25 \text{ mole} = \frac{1}{4} \text{ mole}$.
Number of moles $= \frac{\text{Given volume}}{\text{Molar volume at } STP} = \frac{5.6 \ L}{22.4 \ L/mol} = 0.25 \text{ mole} = \frac{1}{4} \text{ mole}$.
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155
ChemistryMediumMCQMHT CET · 2022
Which of the following species has the highest mass?
A
$1 \ g$ atom of carbon
B
$3.011 \times 10^{23}$ atoms of oxygen
C
$10 \ mL$ of water at room temperature
D
$\frac{1}{2}$ mole of $CH_4$
Solution
(A) Mass of $1 \ g$ atom of carbon $= 12 \ g$.
Mass of $3.011 \times 10^{23}$ atoms of oxygen $= 0.5 \ mol \times 16 \ g/mol = 8 \ g$.
Mass of $10 \ mL$ of water $= 10 \ g$ (since density $\approx 1 \ g/mL$).
Mass of $\frac{1}{2} \ mol$ of $CH_4 = 0.5 \ mol \times 16 \ g/mol = 8 \ g$.
Comparing the masses: $12 \ g > 10 \ g > 8 \ g$.
Therefore,$1 \ g$ atom of carbon has the highest mass.
Mass of $3.011 \times 10^{23}$ atoms of oxygen $= 0.5 \ mol \times 16 \ g/mol = 8 \ g$.
Mass of $10 \ mL$ of water $= 10 \ g$ (since density $\approx 1 \ g/mL$).
Mass of $\frac{1}{2} \ mol$ of $CH_4 = 0.5 \ mol \times 16 \ g/mol = 8 \ g$.
Comparing the masses: $12 \ g > 10 \ g > 8 \ g$.
Therefore,$1 \ g$ atom of carbon has the highest mass.
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156
ChemistryMediumMCQMHT CET · 2022
What is the volume occupied by $32 \ g$ methane gas at $STP$ (in $dm^3$)?
A
$56.0$
B
$33.6$
C
$22.4$
D
$44.8$
Solution
(D) The molar mass of methane $(CH_4)$ is $12 + (4 \times 1) = 16 \ g/mol$.
Number of moles of $CH_4 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{32 \ g}{16 \ g/mol} = 2 \ mol$.
At $STP$,$1 \ mol$ of any ideal gas occupies $22.4 \ dm^3$ of volume.
Therefore,the volume occupied by $2 \ mol$ of $CH_4 = 2 \times 22.4 \ dm^3 = 44.8 \ dm^3$.
Number of moles of $CH_4 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{32 \ g}{16 \ g/mol} = 2 \ mol$.
At $STP$,$1 \ mol$ of any ideal gas occupies $22.4 \ dm^3$ of volume.
Therefore,the volume occupied by $2 \ mol$ of $CH_4 = 2 \times 22.4 \ dm^3 = 44.8 \ dm^3$.
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157
ChemistryMediumMCQMHT CET · 2022
What is the volume occupied by $2 \ g$ of helium gas at $STP$ (in $dm^3$)? (molar mass of helium gas $= 4 \ g \ mol^{-1}$)
A
$5.6$
B
$22.4$
C
$11.2$
D
$2.0$
Solution
(C) The number of moles of $He$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \ g}{4 \ g \ mol^{-1}} = 0.5 \ mol$.
At $STP$,$1 \ mol$ of any ideal gas occupies $22.4 \ dm^3$.
Therefore,the volume occupied by $0.5 \ mol$ of $He$ gas is: $V = 0.5 \ mol \times 22.4 \ dm^3 \ mol^{-1} = 11.2 \ dm^3$.
At $STP$,$1 \ mol$ of any ideal gas occupies $22.4 \ dm^3$.
Therefore,the volume occupied by $0.5 \ mol$ of $He$ gas is: $V = 0.5 \ mol \times 22.4 \ dm^3 \ mol^{-1} = 11.2 \ dm^3$.
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158
ChemistryEasyMCQMHT CET · 2022
What is the number of $CO_2$ molecules in $45.42 \ L$ at $\text{STP}$? (Consider molar volume of gas at $\text{STP} = 22.71 \ L \ mol^{-1}$)
A
$0.913 \times 10^{24}$
B
$1.806 \times 10^{24}$
C
$1.501 \times 10^{24}$
D
$1.204 \times 10^{24}$
Solution
(D) Step $1$: Calculate the number of moles of $CO_2$ using the given molar volume at $\text{STP}$.
Number of moles $= \frac{\text{Given Volume}}{\text{Molar Volume}} = \frac{45.42 \ L}{22.71 \ L \ mol^{-1}} = 2 \ mol$.
Step $2$: Calculate the number of molecules using Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$.
Number of molecules $= \text{moles} \times N_A = 2 \ mol \times 6.022 \times 10^{23} \ mol^{-1} = 1.2044 \times 10^{24}$ molecules.
Number of moles $= \frac{\text{Given Volume}}{\text{Molar Volume}} = \frac{45.42 \ L}{22.71 \ L \ mol^{-1}} = 2 \ mol$.
Step $2$: Calculate the number of molecules using Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$.
Number of molecules $= \text{moles} \times N_A = 2 \ mol \times 6.022 \times 10^{23} \ mol^{-1} = 1.2044 \times 10^{24}$ molecules.
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159
ChemistryEasyMCQMHT CET · 2022
Find the number of hydrogen atoms present in $6.0 \ g$ of urea $(NH_2CONH_2)$:
A
$2.4 \times 10^{23}$
B
$4.06 \times 10^{23}$
C
$2.16 \times 10^{23}$
D
$3.01 \times 10^{23}$
Solution
(A) The molar mass of urea $(NH_2CONH_2)$ is $14 + 2(1) + 12 + 16 + 14 + 2(1) = 60 \ g/mol$.
Moles of urea $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{6.0 \ g}{60 \ g/mol} = 0.1 \ mol$.
Each molecule of urea contains $4$ hydrogen atoms.
Therefore,moles of $H$-atoms $= 0.1 \ mol \times 4 = 0.4 \ mol$.
Number of hydrogen atoms $= \text{moles} \times N_A = 0.4 \times 6.022 \times 10^{23} = 2.4088 \times 10^{23} \approx 2.4 \times 10^{23}$.
Moles of urea $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{6.0 \ g}{60 \ g/mol} = 0.1 \ mol$.
Each molecule of urea contains $4$ hydrogen atoms.
Therefore,moles of $H$-atoms $= 0.1 \ mol \times 4 = 0.4 \ mol$.
Number of hydrogen atoms $= \text{moles} \times N_A = 0.4 \times 6.022 \times 10^{23} = 2.4088 \times 10^{23} \approx 2.4 \times 10^{23}$.
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160
ChemistryMediumMCQMHT CET · 2022
What is the amount of water formed by combustion of $1.6 \ g$ methane (in $g$)?
A
$6.2$
B
$3.2$
C
$3.6$
D
$16$
Solution
(C) The balanced chemical equation for the combustion of methane is: $CH_4 + 2 O_2 \rightarrow CO_2 + 2 H_2 O$
Calculate the moles of methane $(CH_4)$: Molar mass of $CH_4 = 12 + (4 \times 1) = 16 \ g/mol$. Moles $= \frac{1.6 \ g}{16 \ g/mol} = 0.1 \ mol$.
From the stoichiometry of the reaction,$1 \ mol$ of $CH_4$ produces $2 \ mol$ of $H_2 O$. Therefore,$0.1 \ mol$ of $CH_4$ will produce $0.1 \times 2 = 0.2 \ mol$ of $H_2 O$.
Calculate the mass of water $(H_2 O)$: Molar mass of $H_2 O = (2 \times 1) + 16 = 18 \ g/mol$. Mass $= 0.2 \ mol \times 18 \ g/mol = 3.6 \ g$.
Calculate the moles of methane $(CH_4)$: Molar mass of $CH_4 = 12 + (4 \times 1) = 16 \ g/mol$. Moles $= \frac{1.6 \ g}{16 \ g/mol} = 0.1 \ mol$.
From the stoichiometry of the reaction,$1 \ mol$ of $CH_4$ produces $2 \ mol$ of $H_2 O$. Therefore,$0.1 \ mol$ of $CH_4$ will produce $0.1 \times 2 = 0.2 \ mol$ of $H_2 O$.
Calculate the mass of water $(H_2 O)$: Molar mass of $H_2 O = (2 \times 1) + 16 = 18 \ g/mol$. Mass $= 0.2 \ mol \times 18 \ g/mol = 3.6 \ g$.
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161
ChemistryMediumMCQMHT CET · 2022
What quantity of carbon dioxide is produced when $2 \,mol$ of carbon is burnt in $16 \,g$ of dioxygen (in $\,g$)?
A
$44$
B
$88$
C
$11$
D
$22$
Solution
(D) The balanced chemical equation for the combustion of carbon is: $C(s) + O_2(g) \rightarrow CO_2(g)$.
Given amount of carbon = $2 \,mol$.
Given mass of dioxygen $(O_2)$ = $16 \,g$.
Molar mass of $O_2$ = $32 \,g/mol$.
Number of moles of $O_2$ = $\frac{16 \,g}{32 \,g/mol} = 0.5 \,mol$.
According to the stoichiometry,$1 \,mol$ of $C$ reacts with $1 \,mol$ of $O_2$ to produce $1 \,mol$ of $CO_2$.
Since we have $2 \,mol$ of $C$ and only $0.5 \,mol$ of $O_2$,$O_2$ is the Limiting Reagent $(LR)$.
The amount of $CO_2$ produced depends on the amount of $LR$ $(O_2)$.
Thus,$0.5 \,mol$ of $O_2$ will produce $0.5 \,mol$ of $CO_2$.
Mass of $CO_2$ produced = $0.5 \,mol \times 44 \,g/mol = 22 \,g$.
Given amount of carbon = $2 \,mol$.
Given mass of dioxygen $(O_2)$ = $16 \,g$.
Molar mass of $O_2$ = $32 \,g/mol$.
Number of moles of $O_2$ = $\frac{16 \,g}{32 \,g/mol} = 0.5 \,mol$.
According to the stoichiometry,$1 \,mol$ of $C$ reacts with $1 \,mol$ of $O_2$ to produce $1 \,mol$ of $CO_2$.
Since we have $2 \,mol$ of $C$ and only $0.5 \,mol$ of $O_2$,$O_2$ is the Limiting Reagent $(LR)$.
The amount of $CO_2$ produced depends on the amount of $LR$ $(O_2)$.
Thus,$0.5 \,mol$ of $O_2$ will produce $0.5 \,mol$ of $CO_2$.
Mass of $CO_2$ produced = $0.5 \,mol \times 44 \,g/mol = 22 \,g$.
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162
ChemistryMediumMCQMHT CET · 2022
What is the value of percent atom economy if the molar mass of reactants is $36 \ u$ and the mass of the desired product is $27 \ u$ (in $\%$)?
A
$75$
B
$45$
C
$25$
D
$50$
Solution
(A) $\text{Atom economy} = \frac{\text{Mass of desired product}}{\text{Molar mass of all reactants}} \times 100$
$\text{Atom economy} = \frac{27}{36} \times 100 = 75 \%$
$\text{Atom economy} = \frac{27}{36} \times 100 = 75 \%$
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163
ChemistryEasyMCQMHT CET · 2022
Find the quantity of dihydrogen required to prepare $2 \ L$ ammonia gas from $1 \ L$ dinitrogen.
A
$3/2 \ L$
B
$1 \ L$
C
$2 \ L$
D
$3 \ L$
Solution
(D) The balanced chemical equation for the synthesis of ammonia is:
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
According to Gay-Lussac's Law of Gaseous Volumes,gases react in simple ratios by volume at constant temperature and pressure.
From the stoichiometry of the reaction,$1 \ volume$ of $N_2$ reacts with $3 \ volumes$ of $H_2$ to produce $2 \ volumes$ of $NH_3$.
Therefore,to prepare $2 \ L$ of $NH_3$,we require $3 \ L$ of $H_2$ gas.
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
According to Gay-Lussac's Law of Gaseous Volumes,gases react in simple ratios by volume at constant temperature and pressure.
From the stoichiometry of the reaction,$1 \ volume$ of $N_2$ reacts with $3 \ volumes$ of $H_2$ to produce $2 \ volumes$ of $NH_3$.
Therefore,to prepare $2 \ L$ of $NH_3$,we require $3 \ L$ of $H_2$ gas.
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164
ChemistryEasyMCQMHT CET · 2022
What is the quantity of water produced when $34.2 \ g$ of cane sugar is treated with concentrated $H_2SO_4$ under normal conditions of temperature and pressure (in $g$)?
A
$11.0$
B
$19.8$
C
$39.6$
D
$34.2$
Solution
(B) The dehydration reaction of cane sugar (sucrose) with concentrated $H_2SO_4$ is given by:
$C_{12}H_{22}O_{11} \xrightarrow{\text{Conc. } H_2SO_4} 12C + 11H_2O$
From the stoichiometry,$1 \ mol$ of $C_{12}H_{22}O_{11}$ yields $11 \ mol$ of $H_2O$.
The molar mass of $C_{12}H_{22}O_{11} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 342 \ g/mol$.
Number of moles of sugar $= \frac{34.2 \ g}{342 \ g/mol} = 0.1 \ mol$.
Since $1 \ mol$ of sugar produces $11 \ mol$ of $H_2O$,then $0.1 \ mol$ of sugar produces $0.1 \times 11 = 1.1 \ mol$ of $H_2O$.
The mass of $1.1 \ mol$ of $H_2O = 1.1 \ mol \times 18 \ g/mol = 19.8 \ g$.
$C_{12}H_{22}O_{11} \xrightarrow{\text{Conc. } H_2SO_4} 12C + 11H_2O$
From the stoichiometry,$1 \ mol$ of $C_{12}H_{22}O_{11}$ yields $11 \ mol$ of $H_2O$.
The molar mass of $C_{12}H_{22}O_{11} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 342 \ g/mol$.
Number of moles of sugar $= \frac{34.2 \ g}{342 \ g/mol} = 0.1 \ mol$.
Since $1 \ mol$ of sugar produces $11 \ mol$ of $H_2O$,then $0.1 \ mol$ of sugar produces $0.1 \times 11 = 1.1 \ mol$ of $H_2O$.
The mass of $1.1 \ mol$ of $H_2O = 1.1 \ mol \times 18 \ g/mol = 19.8 \ g$.
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165
ChemistryEasyMCQMHT CET · 2022
What is the quantity of sugar charcoal remaining behind after charring $17.1 \ g$ of sugar using conc. $H_2SO_4$ under ideal conditions (in $g$)?
A
$7.2$
B
$14.4$
C
$10.5$
D
$11.4$
Solution
(A) The dehydration reaction of sugar $(C_{12}H_{22}O_{11})$ by conc. $H_2SO_4$ is given by:
$C_{12}H_{22}O_{11} \rightarrow 12C + 11H_2O$
First,calculate the number of moles of sugar:
Molar mass of $C_{12}H_{22}O_{11} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 342 \ g/mol$.
Moles of sugar $= \frac{17.1 \ g}{342 \ g/mol} = 0.05 \ mol$.
According to the stoichiometry,$1 \ mol$ of sugar produces $12 \ mol$ of carbon (charcoal).
Therefore,$0.05 \ mol$ of sugar produces $0.05 \times 12 = 0.6 \ mol$ of carbon.
Mass of charcoal $= 0.6 \ mol \times 12 \ g/mol = 7.2 \ g$.
$C_{12}H_{22}O_{11} \rightarrow 12C + 11H_2O$
First,calculate the number of moles of sugar:
Molar mass of $C_{12}H_{22}O_{11} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 342 \ g/mol$.
Moles of sugar $= \frac{17.1 \ g}{342 \ g/mol} = 0.05 \ mol$.
According to the stoichiometry,$1 \ mol$ of sugar produces $12 \ mol$ of carbon (charcoal).
Therefore,$0.05 \ mol$ of sugar produces $0.05 \times 12 = 0.6 \ mol$ of carbon.
Mass of charcoal $= 0.6 \ mol \times 12 \ g/mol = 7.2 \ g$.
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166
ChemistryMediumMCQMHT CET · 2022
What is the density of water in $kg \ dm^{-3}$ if its density in $g \ cm^{-3}$ is $0.863$?
A
$4.60$
B
$7.86$
C
$8.63$
D
$0.863$
Solution
(D) The density is given as $0.863 \ g \ cm^{-3}$.
To convert $g \ cm^{-3}$ to $kg \ dm^{-3}$:
$1 \ g = 10^{-3} \ kg$
$1 \ cm^{3} = (10^{-1} \ dm)^{3} = 10^{-3} \ dm^{3}$
Therefore,$0.863 \ g \ cm^{-3} = \frac{0.863 \times 10^{-3} \ kg}{10^{-3} \ dm^{3}} = 0.863 \ kg \ dm^{-3}$.
To convert $g \ cm^{-3}$ to $kg \ dm^{-3}$:
$1 \ g = 10^{-3} \ kg$
$1 \ cm^{3} = (10^{-1} \ dm)^{3} = 10^{-3} \ dm^{3}$
Therefore,$0.863 \ g \ cm^{-3} = \frac{0.863 \times 10^{-3} \ kg}{10^{-3} \ dm^{3}} = 0.863 \ kg \ dm^{-3}$.
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167
ChemistryEasyMCQMHT CET · 2022
Find the formula weight of the reactant if the formula weight of the product is $54 \ u$ and the percent atom economy is $75$. (in $u$)
A
$72$
B
$24$
C
$30$
D
$80$
Solution
(A) The formula for atom economy is given by: $\text{Atom economy} = \frac{\text{Formula weight of desired product}}{\text{Formula weight of all reactants}} \times 100$
Given: $\text{Atom economy} = 75$,$\text{Product weight} = 54 \ u$.
Let the reactant weight be $x$.
$75 = \frac{54}{x} \times 100$
$x = \frac{54 \times 100}{75}$
$x = \frac{5400}{75} = 72 \ u$
Therefore,the formula weight of the reactant is $72 \ u$.
Given: $\text{Atom economy} = 75$,$\text{Product weight} = 54 \ u$.
Let the reactant weight be $x$.
$75 = \frac{54}{x} \times 100$
$x = \frac{54 \times 100}{75}$
$x = \frac{5400}{75} = 72 \ u$
Therefore,the formula weight of the reactant is $72 \ u$.
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168
ChemistryEasyMCQMHT CET · 2022
What is the percent atom economy if the formula weight of reactants and the formula weight of products are $45 \ u$ and $35 \ u$ respectively (in $\%$)?
A
$77.8$
B
$90.0$
C
$80.5$
D
$71.0$
Solution
(A) The formula for percent atom economy is:
$\text{Percent atom economy} = \frac{\text{Formula weight of desired product}}{\text{Formula weight of all reactants}} \times 100$
Given:
$\text{Formula weight of reactants} = 45 \ u$
$\text{Formula weight of products} = 35 \ u$
Calculation:
$\text{Percent atom economy} = \frac{35}{45} \times 100 = 77.77... \% \approx 77.8 \%$
$\text{Percent atom economy} = \frac{\text{Formula weight of desired product}}{\text{Formula weight of all reactants}} \times 100$
Given:
$\text{Formula weight of reactants} = 45 \ u$
$\text{Formula weight of products} = 35 \ u$
Calculation:
$\text{Percent atom economy} = \frac{35}{45} \times 100 = 77.77... \% \approx 77.8 \%$
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169
ChemistryEasyMCQMHT CET · 2022
Calculate the final volume of a gas when the pressure of $60 \ mL$ of gas is increased from $1 \ atm$ to $1.5 \ atm$,keeping the temperature constant.
A
$5 \times 10^{-2} \ dm^3$
B
$2 \times 10^{-2} \ dm^3$
C
$3 \times 10^{-2} \ dm^3$
D
$4 \times 10^{-2} \ dm^3$
Solution
(D) According to Boyle's Law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 1 \ atm$,$V_1 = 60 \ mL$,$P_2 = 1.5 \ atm$.
Substituting the values: $1 \ atm \times 60 \ mL = 1.5 \ atm \times V_2$.
$V_2 = \frac{60}{1.5} \ mL = 40 \ mL$.
Since $1 \ dm^3 = 1000 \ mL$,$40 \ mL = 40 \times 10^{-3} \ dm^3 = 4 \times 10^{-2} \ dm^3$.
Given: $P_1 = 1 \ atm$,$V_1 = 60 \ mL$,$P_2 = 1.5 \ atm$.
Substituting the values: $1 \ atm \times 60 \ mL = 1.5 \ atm \times V_2$.
$V_2 = \frac{60}{1.5} \ mL = 40 \ mL$.
Since $1 \ dm^3 = 1000 \ mL$,$40 \ mL = 40 \times 10^{-3} \ dm^3 = 4 \times 10^{-2} \ dm^3$.
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170
ChemistryEasyMCQMHT CET · 2022
Calculate the pressure of $1.5 \ mol$ of gas having volume $3 \ dm^3$ at $300 \ K$ $(R=0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1})$. (in $atm$)
A
$12.32$
B
$14.6$
C
$10.25$
D
$15.3$
Solution
(A) Using the ideal gas equation: $PV = nRT$
Given:
$n = 1.5 \ mol$
$V = 3 \ dm^3$
$T = 300 \ K$
$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$
Substituting the values:
$P \times 3 = 1.5 \times 0.0821 \times 300$
$P \times 3 = 36.945$
$P = \frac{36.945}{3} = 12.315 \ atm \approx 12.32 \ atm$.
Given:
$n = 1.5 \ mol$
$V = 3 \ dm^3$
$T = 300 \ K$
$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$
Substituting the values:
$P \times 3 = 1.5 \times 0.0821 \times 300$
$P \times 3 = 36.945$
$P = \frac{36.945}{3} = 12.315 \ atm \approx 12.32 \ atm$.
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171
ChemistryEasyMCQMHT CET · 2022
Calculate the number of moles of the gas having volume $2.5 \ L$ at $300 \ K$ and $4.5 \ atm$? $(R=0.0821 \ atm \ dm^3 \ K^{-1} \ mol^{-1})$
A
$0.62$
B
$0.46$
C
$0.56$
D
$0.70$
Solution
(B) Using the ideal gas equation: $PV = nRT$
Given: $P = 4.5 \ atm$,$V = 2.5 \ L$,$T = 300 \ K$,and $R = 0.0821 \ atm \ L \ K^{-1} \ mol^{-1}$.
Substituting the values: $4.5 \times 2.5 = n \times 0.0821 \times 300$
$11.25 = n \times 24.63$
$n = \frac{11.25}{24.63} \approx 0.4567 \ mol$
Rounding to two decimal places,we get $n \approx 0.46 \ mol$.
Given: $P = 4.5 \ atm$,$V = 2.5 \ L$,$T = 300 \ K$,and $R = 0.0821 \ atm \ L \ K^{-1} \ mol^{-1}$.
Substituting the values: $4.5 \times 2.5 = n \times 0.0821 \times 300$
$11.25 = n \times 24.63$
$n = \frac{11.25}{24.63} \approx 0.4567 \ mol$
Rounding to two decimal places,we get $n \approx 0.46 \ mol$.
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172
ChemistryEasyMCQMHT CET · 2022
Which of the following is a correct relation for Gay-Lussac's law?
A
$P \propto T$ (at constant volume)
B
$V \propto n$ (at constant temperature and pressure)
C
$P \propto \frac{1}{V}$ (at constant temperature)
D
$V \propto T$ (at constant pressure)
Solution
(A) Gay-Lussac's law states that the pressure of a given mass of gas is directly proportional to its absolute temperature,provided the volume remains constant.
Mathematically,this is expressed as $P \propto T$ at constant volume and mass.
Mathematically,this is expressed as $P \propto T$ at constant volume and mass.
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173
ChemistryMediumMCQMHT CET · 2022
If the mixture of $7 \ g$ of $N_2$ and $8 \ g$ of $Ar$ in a cylinder has a total pressure of $27 \ bar$,what is the partial pressure of $N_2$ (in $bar$)? (Atomic mass of $N = 14 \ g \ mol^{-1}$,$Ar = 40 \ g \ mol^{-1}$)
A
$18$
B
$12$
C
$15$
D
$9$
Solution
(C) Moles of $N_2 = \frac{7}{28} = 0.25 \ mol$
Moles of $Ar = \frac{8}{40} = 0.20 \ mol$
Total moles $= 0.25 + 0.20 = 0.45 \ mol$
Mole fraction of $N_2$ $(X_{N_2})$ $= \frac{0.25}{0.45} = \frac{5}{9}$
Partial pressure of $N_2 = X_{N_2} \times P_{total}$
Partial pressure of $N_2 = \frac{5}{9} \times 27 \ bar = 15 \ bar$
Moles of $Ar = \frac{8}{40} = 0.20 \ mol$
Total moles $= 0.25 + 0.20 = 0.45 \ mol$
Mole fraction of $N_2$ $(X_{N_2})$ $= \frac{0.25}{0.45} = \frac{5}{9}$
Partial pressure of $N_2 = X_{N_2} \times P_{total}$
Partial pressure of $N_2 = \frac{5}{9} \times 27 \ bar = 15 \ bar$
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174
ChemistryMediumMCQMHT CET · 2022
Which of the following gases is most difficult to liquefy?
A
$O_2$
B
$SO_2$
C
$Cl_2$
D
$NH_3$
Solution
(A) The ease of liquefaction of a gas depends on the magnitude of intermolecular forces of attraction.
$SO_2$,$Cl_2$,and $NH_3$ are polar molecules,which possess stronger dipole-dipole interactions.
$O_2$ is a non-polar molecule,which only possesses weak London dispersion forces.
Therefore,$O_2$ has the lowest critical temperature and is the most difficult to liquefy.
$SO_2$,$Cl_2$,and $NH_3$ are polar molecules,which possess stronger dipole-dipole interactions.
$O_2$ is a non-polar molecule,which only possesses weak London dispersion forces.
Therefore,$O_2$ has the lowest critical temperature and is the most difficult to liquefy.
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175
ChemistryEasyMCQMHT CET · 2022
What is the number of nucleons present in an atom having $29$ electrons and $34$ neutrons in it?
A
$29$
B
$34$
C
$05$
D
$63$
Solution
(D) In a neutral atom,the number of electrons is equal to the number of protons.
Given that the atom has $29$ electrons,it must have $29$ protons.
The number of nucleons is the sum of the number of protons and the number of neutrons.
$\text{Number of nucleons} = \text{Number of protons} + \text{Number of neutrons} = 29 + 34 = 63$.
Given that the atom has $29$ electrons,it must have $29$ protons.
The number of nucleons is the sum of the number of protons and the number of neutrons.
$\text{Number of nucleons} = \text{Number of protons} + \text{Number of neutrons} = 29 + 34 = 63$.
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176
ChemistryMediumMCQMHT CET · 2022
Calculate the wave number of a photon emitted during the transition from the orbit $n = 2$ to $n = 1$ in a hydrogen atom $(R_{H} = 109677 \ cm^{-1})$. (in $cm^{-1}$)
A
$72740$
B
$92820$
C
$82258$
D
$83560$
Solution
(C) The wave number $\bar{\nu}$ is calculated using the Rydberg formula: $\bar{\nu} = R_{H} (\frac{1}{n_1^2} - \frac{1}{n_2^2})$
Given $n_1 = 1$,$n_2 = 2$,and $R_{H} = 109677 \ cm^{-1}$.
Substituting the values: $\bar{\nu} = 109677 \times (\frac{1}{1^2} - \frac{1}{2^2})$
$\bar{\nu} = 109677 \times (1 - 0.25) = 109677 \times 0.75$
$\bar{\nu} = 82257.75 \ cm^{-1} \approx 82258 \ cm^{-1}$.
Given $n_1 = 1$,$n_2 = 2$,and $R_{H} = 109677 \ cm^{-1}$.
Substituting the values: $\bar{\nu} = 109677 \times (\frac{1}{1^2} - \frac{1}{2^2})$
$\bar{\nu} = 109677 \times (1 - 0.25) = 109677 \times 0.75$
$\bar{\nu} = 82257.75 \ cm^{-1} \approx 82258 \ cm^{-1}$.
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177
ChemistryEasyMCQMHT CET · 2022
What is the energy associated with the first orbit of $Li^{2+}$ $(R_{H} = 2.18 \times 10^{-18} \ J)$?
A
$-8.72 \times 10^{-18} \ J$
B
$-34.88 \times 10^{-18} \ J$
C
$-2.18 \times 10^{-18} \ J$
D
$-19.62 \times 10^{-18} \ J$
Solution
(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -R_H \times \frac{Z^2}{n^2}$.
For $Li^{2+}$,the atomic number $Z = 3$ and for the first orbit,$n = 1$.
Substituting the values: $E_1 = -2.18 \times 10^{-18} \times \frac{3^2}{1^2} \ J$.
$E_1 = -2.18 \times 10^{-18} \times 9 \ J$.
$E_1 = -19.62 \times 10^{-18} \ J$.
For $Li^{2+}$,the atomic number $Z = 3$ and for the first orbit,$n = 1$.
Substituting the values: $E_1 = -2.18 \times 10^{-18} \times \frac{3^2}{1^2} \ J$.
$E_1 = -2.18 \times 10^{-18} \times 9 \ J$.
$E_1 = -19.62 \times 10^{-18} \ J$.
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178
ChemistryEasyMCQMHT CET · 2022
Calculate the radius of the first orbit of $Be^{3+}$. (in $pm$)
A
$13.23$
B
$52.9$
C
$17.63$
D
$13.25$
Solution
(A) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For $Be^{3+}$, the atomic number $Z = 4$ and for the first orbit, $n = 1$.
Substituting these values: $r_1 = 0.529 \times \frac{1^2}{4} \ \mathring{A}$.
$r_1 = 0.529 \times 0.25 \ \mathring{A} = 0.13225 \ \mathring{A}$.
Since $1 \ \mathring{A} = 100 \ pm$, we have $r_1 = 0.13225 \times 100 \ pm = 13.225 \ pm$, which is approximately $13.23 \ pm$.
For $Be^{3+}$, the atomic number $Z = 4$ and for the first orbit, $n = 1$.
Substituting these values: $r_1 = 0.529 \times \frac{1^2}{4} \ \mathring{A}$.
$r_1 = 0.529 \times 0.25 \ \mathring{A} = 0.13225 \ \mathring{A}$.
Since $1 \ \mathring{A} = 100 \ pm$, we have $r_1 = 0.13225 \times 100 \ pm = 13.225 \ pm$, which is approximately $13.23 \ pm$.
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179
ChemistryEasyMCQMHT CET · 2022
Which of the following is not a hydrogen-like species?
A
$He$
B
$Be^{3+}$
C
$He^{+}$
D
$Li^{2+}$
Solution
(A) hydrogen-like species is an ion that contains only one electron.
$He$ (Helium) has an atomic number of $2$ and its electronic configuration is $1s^2$,meaning it has $2$ electrons.
$He^{+}$ has $1$ electron.
$Li^{2+}$ has $1$ electron.
$Be^{3+}$ has $1$ electron.
Therefore,$He$ is not a hydrogen-like species.
$He$ (Helium) has an atomic number of $2$ and its electronic configuration is $1s^2$,meaning it has $2$ electrons.
$He^{+}$ has $1$ electron.
$Li^{2+}$ has $1$ electron.
$Be^{3+}$ has $1$ electron.
Therefore,$He$ is not a hydrogen-like species.
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180
ChemistryEasyMCQMHT CET · 2022
What is the relation between radius, order of orbit, and nuclear charge for hydrogen-like species?
A
$r_n = \frac{a_0 n^2}{z} \text{ pm}$
B
$r_n = \frac{a_0 n^2}{z^2} \text{ pm}$
C
$r_n = \frac{a_0 n}{n^2} \text{ pm}$
D
$r_n = \frac{a_0 z}{n^2} \text{ pm}$
Solution
$(A)$ The radius of the $n^{th}$ orbit for a hydrogen-like species is given by the formula:
$r_n = a_0 \frac{n^2}{z}$
where $a_0$ is the Bohr radius $(0.529 \ \mathring{A})$, $n$ is the principal quantum number (order of orbit), and $z$ is the atomic number (nuclear charge).
$r_n = a_0 \frac{n^2}{z}$
where $a_0$ is the Bohr radius $(0.529 \ \mathring{A})$, $n$ is the principal quantum number (order of orbit), and $z$ is the atomic number (nuclear charge).
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181
ChemistryEasyMCQMHT CET · 2022
What does magnetic quantum number describe?
A
Orientation of an orbital in the given subshell
B
Spin of an electron
C
Shape of an orbital
D
Size of an orbital
Solution
(A) The magnetic quantum number $(m_l)$ describes the orientation of an orbital in $3D$ space relative to the chosen coordinate system.
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182
ChemistryEasyMCQMHT CET · 2022
Which among the following elements develops noble gas configuration in $+1$ state?
A
$Fr$
B
$Ca$
C
$Mg$
D
$Sr$
Solution
(A) The electronic configuration of Francium $(Fr)$ is $[Rn] 7s^1$.
When it loses one electron to form a $+1$ oxidation state,it becomes $Fr^+ = [Rn]$.
Since $[Rn]$ is the electronic configuration of the noble gas Radon,$Fr$ achieves a noble gas configuration in the $+1$ state.
Other elements like $Ca$,$Mg$,and $Sr$ are alkaline earth metals and typically form $+2$ oxidation states to achieve noble gas configurations.
When it loses one electron to form a $+1$ oxidation state,it becomes $Fr^+ = [Rn]$.
Since $[Rn]$ is the electronic configuration of the noble gas Radon,$Fr$ achieves a noble gas configuration in the $+1$ state.
Other elements like $Ca$,$Mg$,and $Sr$ are alkaline earth metals and typically form $+2$ oxidation states to achieve noble gas configurations.
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183
ChemistryMCQMHT CET · 2022
Which of the following is $NOT$ a characteristic of chemisorption?
A
Heat of adsorption is in the range of $40-200 \ kJ \ mol^{-1}$
B
It is specific.
C
It is irreversible.
D
Formation of multimolecular layer of adsorbate.
Solution
(D) Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent.
Due to the formation of chemical bonds,it is highly specific and irreversible in nature.
The heat of adsorption for chemisorption is high,typically in the range of $40-200 \ kJ \ mol^{-1}$.
However,chemisorption results in the formation of a unimolecular layer,not a multimolecular layer.
Multimolecular layer formation is a characteristic of physisorption.
Due to the formation of chemical bonds,it is highly specific and irreversible in nature.
The heat of adsorption for chemisorption is high,typically in the range of $40-200 \ kJ \ mol^{-1}$.
However,chemisorption results in the formation of a unimolecular layer,not a multimolecular layer.
Multimolecular layer formation is a characteristic of physisorption.
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184
ChemistryMCQMHT CET · 2022
Which is $NOT$ an example of a macromolecular colloid?
A
Polythene
B
Soap
C
Nylon
D
Proteins
Solution
(B) Macromolecular colloids are formed by large molecules like polymers (e.g.,$Polythene$,$Nylon$,$Proteins$,$Starch$,$Cellulose$).
Soap is an example of an associated colloid (micelle),not a macromolecular colloid.
Therefore,the correct answer is $Soap$.
Soap is an example of an associated colloid (micelle),not a macromolecular colloid.
Therefore,the correct answer is $Soap$.
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185
ChemistryEasyMCQMHT CET · 2022
Which among the following pairs represents extensive and intensive properties respectively?
A
Volume and number of moles
B
Volume and pressure
C
Surface tension and heat capacity
D
Internal energy and temperature
Solution
(B) Extensive properties depend on the amount of matter present in the system (e.g.,$Volume$,$Number \ of \ moles$,$Internal \ energy$,$Heat \ capacity$).
Intensive properties are independent of the amount of matter present in the system (e.g.,$Pressure$,$Temperature$,$Surface \ tension$,$Density$).
In option $B$,$Volume$ is an extensive property and $Pressure$ is an intensive property. Therefore,the pair ($Volume$,$Pressure$) matches the requirement.
Intensive properties are independent of the amount of matter present in the system (e.g.,$Pressure$,$Temperature$,$Surface \ tension$,$Density$).
In option $B$,$Volume$ is an extensive property and $Pressure$ is an intensive property. Therefore,the pair ($Volume$,$Pressure$) matches the requirement.
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186
ChemistryMediumMCQMHT CET · 2022
Which among the following properties is $NOT$ a state function?
A
Enthalpy
B
Volume
C
Pressure
D
Work
Solution
(D) state function is a property whose value depends only on the current state of the system and not on the path taken to reach that state. $Enthalpy$ $(H)$,$Volume$ $(V)$,and $Pressure$ $(P)$ are state functions. $Work$ $(w)$ and $Heat$ $(q)$ are path functions,as their values depend on the process or path followed. Therefore,$Work$ is not a state function.
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187
ChemistryEasyMCQMHT CET · 2022
Which among the following is $NOT$ an intensive property?
A
Heat capacity
B
Viscosity
C
Pressure
D
Surface tension
Solution
(A) An intensive property is independent of the amount of matter present in the system.
An extensive property depends on the amount of matter present in the system.
Heat capacity $(C)$ is defined as the product of mass $(m)$ and specific heat capacity $(c)$,i.e.,$C = m \times c$.
Since it depends on the mass of the substance,heat capacity is an extensive property.
Viscosity,pressure,and surface tension are independent of the amount of matter and are therefore intensive properties.
An extensive property depends on the amount of matter present in the system.
Heat capacity $(C)$ is defined as the product of mass $(m)$ and specific heat capacity $(c)$,i.e.,$C = m \times c$.
Since it depends on the mass of the substance,heat capacity is an extensive property.
Viscosity,pressure,and surface tension are independent of the amount of matter and are therefore intensive properties.
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188
ChemistryEasyMCQMHT CET · 2022
If $65 \ kJ$ of work is done on the system and it releases $25 \ kJ$ of heat,what is the change in internal energy of the system (in $kJ$)?
A
$16.25$
B
$40$
C
$90$
D
$2.6$
Solution
(B) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = q + w$.
Here,work is done on the system,so $w = +65 \ kJ$.
The system releases heat,so $q = -25 \ kJ$.
Substituting these values: $\Delta U = -25 \ kJ + 65 \ kJ = +40 \ kJ$.
Here,work is done on the system,so $w = +65 \ kJ$.
The system releases heat,so $q = -25 \ kJ$.
Substituting these values: $\Delta U = -25 \ kJ + 65 \ kJ = +40 \ kJ$.
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189
ChemistryMediumMCQMHT CET · 2022
Under isothermal condition,a gas expands from $0.2 \ dm^3$ to $0.8 \ dm^3$ against a constant external pressure of $2 \ bar$ at $300 \ K$. Find the work done by the gas. (in $J$)
A
$160$
B
$-120$
C
$-40$
D
$20$
Solution
(B) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$.
Given: $P_{ext} = 2 \ bar$,$V_1 = 0.2 \ dm^3$,$V_2 = 0.8 \ dm^3$.
Change in volume: $\Delta V = V_2 - V_1 = 0.8 \ dm^3 - 0.2 \ dm^3 = 0.6 \ dm^3$.
Substituting the values: $W = -2 \ bar \times 0.6 \ dm^3 = -1.2 \ bar \cdot dm^3$.
Since $1 \ bar \cdot dm^3 = 100 \ J$,we convert the units: $W = -1.2 \times 100 \ J = -120 \ J$.
Thus,the work done by the gas is $-120 \ J$.
Given: $P_{ext} = 2 \ bar$,$V_1 = 0.2 \ dm^3$,$V_2 = 0.8 \ dm^3$.
Change in volume: $\Delta V = V_2 - V_1 = 0.8 \ dm^3 - 0.2 \ dm^3 = 0.6 \ dm^3$.
Substituting the values: $W = -2 \ bar \times 0.6 \ dm^3 = -1.2 \ bar \cdot dm^3$.
Since $1 \ bar \cdot dm^3 = 100 \ J$,we convert the units: $W = -1.2 \times 100 \ J = -120 \ J$.
Thus,the work done by the gas is $-120 \ J$.
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190
ChemistryMediumMCQMHT CET · 2022
Calculate $\Delta U$ if $2 \ kJ$ heat is released and $10 \ kJ$ of work is done on the system. (in $kJ$)
A
$20$
B
$12$
C
$8$
D
$5$
Solution
(C) According to the $I^{st}$ law of thermodynamics:
$\Delta U = q + w$
Here,heat is released,so $q = -2 \ kJ$.
Work is done on the system,so $w = +10 \ kJ$.
Therefore,$\Delta U = -2 \ kJ + 10 \ kJ = 8 \ kJ$.
$\Delta U = q + w$
Here,heat is released,so $q = -2 \ kJ$.
Work is done on the system,so $w = +10 \ kJ$.
Therefore,$\Delta U = -2 \ kJ + 10 \ kJ = 8 \ kJ$.
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191
ChemistryMediumMCQMHT CET · 2022
An ideal gas is compressed from a volume of $1 \ m^3$ to $0.5 \ m^3$ at a constant pressure of $0.2 \ bar$. What is the value of work done on the gas during compression (in $kJ$)?
A
$10$
B
$201.0$
C
$497$
D
$190$
Solution
(A) The work done during the compression of a gas at constant pressure is given by the formula: $W = -P_{ext} \Delta V$.
Given: $P_{ext} = 0.2 \ bar$,$V_1 = 1 \ m^3$,$V_2 = 0.5 \ m^3$.
$\Delta V = V_2 - V_1 = 0.5 \ m^3 - 1 \ m^3 = -0.5 \ m^3$.
$W = -(0.2 \ bar) \times (-0.5 \ m^3) = 0.1 \ bar \cdot m^3$.
Since $1 \ bar = 10^5 \ Pa$ and $1 \ Pa \cdot m^3 = 1 \ J$,then $1 \ bar \cdot m^3 = 10^5 \ J = 100 \ kJ$.
Therefore,$W = 0.1 \times 100 \ kJ = 10 \ kJ$.
Given: $P_{ext} = 0.2 \ bar$,$V_1 = 1 \ m^3$,$V_2 = 0.5 \ m^3$.
$\Delta V = V_2 - V_1 = 0.5 \ m^3 - 1 \ m^3 = -0.5 \ m^3$.
$W = -(0.2 \ bar) \times (-0.5 \ m^3) = 0.1 \ bar \cdot m^3$.
Since $1 \ bar = 10^5 \ Pa$ and $1 \ Pa \cdot m^3 = 1 \ J$,then $1 \ bar \cdot m^3 = 10^5 \ J = 100 \ kJ$.
Therefore,$W = 0.1 \times 100 \ kJ = 10 \ kJ$.
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192
ChemistryMediumMCQMHT CET · 2022
What is the work done at $300 \ K$ for the reaction?
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)}$ (in $J$)?
(Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)}$ (in $J$)?
(Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)
A
$-1247.1$
B
$-247$
C
$850.7$
D
$1247$
Solution
(D) The work done in a chemical reaction is given by the formula $W = -\Delta n_{g} RT$.
First,calculate the change in the number of moles of gaseous species,$\Delta n_{g} = n_{p(g)} - n_{r(g)}$.
For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)}$,$\Delta n_{g} = 1 - (1 + 0.5) = -0.5 \ mol$.
Now,substitute the values into the work formula: $W = -(-0.5) \times 8.314 \times 300$.
$W = 0.5 \times 8.314 \times 300 = 1247.1 \ J$.
Note: If the question asks for work done $BY$ the system,it is $1247.1 \ J$. If it asks for work done $ON$ the system,it is $-1247.1 \ J$. Given the options,the magnitude is $1247.1 \ J$.
First,calculate the change in the number of moles of gaseous species,$\Delta n_{g} = n_{p(g)} - n_{r(g)}$.
For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)}$,$\Delta n_{g} = 1 - (1 + 0.5) = -0.5 \ mol$.
Now,substitute the values into the work formula: $W = -(-0.5) \times 8.314 \times 300$.
$W = 0.5 \times 8.314 \times 300 = 1247.1 \ J$.
Note: If the question asks for work done $BY$ the system,it is $1247.1 \ J$. If it asks for work done $ON$ the system,it is $-1247.1 \ J$. Given the options,the magnitude is $1247.1 \ J$.
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193
ChemistryMediumMCQMHT CET · 2022
$A$ certain system is exothermic by $260 \ kJ$ and does $10 \ kJ$ of work. What is the change in internal energy (in $kJ$)?
A
$-250$
B
$-540$
C
$-140$
D
$-270$
Solution
(D) According to the first law of thermodynamics,the change in internal energy is given by $\Delta U = Q + W$.
Since the system is exothermic,the heat released is $Q = -260 \ kJ$.
Since the system does work,the work done by the system is $W = -10 \ kJ$.
Therefore,$\Delta U = -260 \ kJ + (-10 \ kJ) = -270 \ kJ$.
Since the system is exothermic,the heat released is $Q = -260 \ kJ$.
Since the system does work,the work done by the system is $W = -10 \ kJ$.
Therefore,$\Delta U = -260 \ kJ + (-10 \ kJ) = -270 \ kJ$.
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194
ChemistryMediumMCQMHT CET · 2022
$A$ gas is allowed to expand in an insulated container against a constant external pressure of $2.5 \ bar$ from $4.5 \ dm^3$ to $7 \times 10^{-3} \ m^3$. What is the change in internal energy of the gas (in $J$)?
A
$-625$
B
$-312.5$
C
$-112.3$
D
$-3.25$
Solution
(A) Since the container is insulated,the process is adiabatic,so $\delta Q = 0$.
According to the first law of thermodynamics,$\Delta U = \delta Q + \delta W$.
Here,$\delta W = -P_{ext} \times \Delta V$.
Given: $P_{ext} = 2.5 \ bar = 2.5 \times 10^5 \ Pa$,$V_1 = 4.5 \ dm^3 = 4.5 \times 10^{-3} \ m^3$,$V_2 = 7 \times 10^{-3} \ m^3$.
$\Delta V = (7 - 4.5) \times 10^{-3} \ m^3 = 2.5 \times 10^{-3} \ m^3$.
$\Delta U = 0 + [-(2.5 \times 10^5 \ Pa) \times (2.5 \times 10^{-3} \ m^3)]$.
$\Delta U = -6.25 \times 10^2 \ J = -625 \ J$.
According to the first law of thermodynamics,$\Delta U = \delta Q + \delta W$.
Here,$\delta W = -P_{ext} \times \Delta V$.
Given: $P_{ext} = 2.5 \ bar = 2.5 \times 10^5 \ Pa$,$V_1 = 4.5 \ dm^3 = 4.5 \times 10^{-3} \ m^3$,$V_2 = 7 \times 10^{-3} \ m^3$.
$\Delta V = (7 - 4.5) \times 10^{-3} \ m^3 = 2.5 \times 10^{-3} \ m^3$.
$\Delta U = 0 + [-(2.5 \times 10^5 \ Pa) \times (2.5 \times 10^{-3} \ m^3)]$.
$\Delta U = -6.25 \times 10^2 \ J = -625 \ J$.
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195
ChemistryMediumMCQMHT CET · 2022
Calculate the change in internal energy of the system if $37.6 \ J$ of work is done by the system with heat loss of $14.6 \ J$ (in $J$)?
A
$-52.2$
B
$-549$
C
$-12.57$
D
$-23.0$
Solution
(A) According to the $I^{st}$ law of thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = q + w$.
Here,work is done by the system,so $w = -37.6 \ J$.
Heat is lost by the system,so $q = -14.6 \ J$.
Substituting these values into the equation:
$\Delta U = -14.6 \ J + (-37.6 \ J) = -52.2 \ J$.
Here,work is done by the system,so $w = -37.6 \ J$.
Heat is lost by the system,so $q = -14.6 \ J$.
Substituting these values into the equation:
$\Delta U = -14.6 \ J + (-37.6 \ J) = -52.2 \ J$.
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196
ChemistryMediumMCQMHT CET · 2022
Calculate the new volume of a gas at constant pressure when temperature is increased to $546 \ K$.
[Initial volume of a gas at $273 \ K$ is $4 \ dm^3$] (in $dm^3$)
[Initial volume of a gas at $273 \ K$ is $4 \ dm^3$] (in $dm^3$)
A
$8$
B
$2$
C
$4$
D
$5$
Solution
(A) According to Charles's Law,at constant pressure,the volume of a fixed mass of gas is directly proportional to its absolute temperature: $V_1 / T_1 = V_2 / T_2$.
Given: $V_1 = 4 \ dm^3$,$T_1 = 273 \ K$,$T_2 = 546 \ K$.
Substituting the values: $4 / 273 = V_2 / 546$.
$V_2 = (4 \times 546) / 273 = 4 \times 2 = 8 \ dm^3$.
Given: $V_1 = 4 \ dm^3$,$T_1 = 273 \ K$,$T_2 = 546 \ K$.
Substituting the values: $4 / 273 = V_2 / 546$.
$V_2 = (4 \times 546) / 273 = 4 \times 2 = 8 \ dm^3$.
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197
ChemistryMediumMCQMHT CET · 2022
What is the change in internal energy if a system does $140 \ kJ$ of work on the surroundings and $40 \ kJ$ of heat is added to the system (in $kJ$)?
A
$-200$
B
$-180$
C
$-100$
D
$-280$
Solution
(C) According to the first law of thermodynamics,the change in internal energy ($\Delta U$ or $\Delta E$) is given by the equation: $\Delta E = q + w$.
Here,heat is added to the system,so $q = +40 \ kJ$.
The system does work on the surroundings,so $w = -140 \ kJ$.
Substituting these values into the equation: $\Delta E = 40 \ kJ + (-140 \ kJ) = -100 \ kJ$.
Here,heat is added to the system,so $q = +40 \ kJ$.
The system does work on the surroundings,so $w = -140 \ kJ$.
Substituting these values into the equation: $\Delta E = 40 \ kJ + (-140 \ kJ) = -100 \ kJ$.
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198
ChemistryMediumMCQMHT CET · 2022
$A$ gas is allowed to expand against a constant external pressure of $2.5 \ bar$ from an initial volume of $2.5 \ L$ to a final volume of $4.5 \ L$. What is the amount of work done (in $J$)?
A
$-375$
B
$-650$
C
$-250$
D
$-500$
Solution
(D) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$
Given:
$P_{ext} = 2.5 \ bar$
$V_1 = 2.5 \ L$
$V_2 = 4.5 \ L$
Calculation:
$\Delta V = V_2 - V_1 = 4.5 \ L - 2.5 \ L = 2.0 \ L$
$W = -2.5 \ bar \times 2.0 \ L = -5.0 \ bar \cdot L$
Since $1 \ bar \cdot L = 100 \ J$:
$W = -5.0 \times 100 \ J = -500 \ J$
Given:
$P_{ext} = 2.5 \ bar$
$V_1 = 2.5 \ L$
$V_2 = 4.5 \ L$
Calculation:
$\Delta V = V_2 - V_1 = 4.5 \ L - 2.5 \ L = 2.0 \ L$
$W = -2.5 \ bar \times 2.0 \ L = -5.0 \ bar \cdot L$
Since $1 \ bar \cdot L = 100 \ J$:
$W = -5.0 \times 100 \ J = -500 \ J$
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199
ChemistryMediumMCQMHT CET · 2022
What is the internal energy change when $2$ moles of an ideal gas at $25^{\circ} C$ is compressed isothermally and reversibly from $1.0 \ bar$ to $10.0 \ bar$?
A
zero $kJ$
B
$34.23 \ kJ$
C
$22.82 \ kJ$
D
$11.41 \ kJ$
Solution
(A) For an ideal gas,the internal energy $(U)$ is a function of temperature only,i.e.,$U = f(T)$.
Since the process is isothermal,the temperature remains constant,so $\Delta T = 0$.
Therefore,the change in internal energy $\Delta U = nC_v\Delta T = 0$.
Since the process is isothermal,the temperature remains constant,so $\Delta T = 0$.
Therefore,the change in internal energy $\Delta U = nC_v\Delta T = 0$.
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200
ChemistryMediumMCQMHT CET · 2022
What is the internal energy change when $X \ J$ of work is done on the system and $Y \ J$ of heat is transferred to the surrounding?
A
$X + Y \ J$
B
$X - Y \ J$
C
$Y - X \ J$
D
$-X - Y \ J$
Solution
(B) According to the first law of thermodynamics,the change in internal energy is given by $\Delta U = q + w$.
When work is done on the system,$w = +X \ J$.
When heat is transferred to the surrounding,$q = -Y \ J$.
Therefore,$\Delta U = (-Y) + (+X) = X - Y \ J$.
When work is done on the system,$w = +X \ J$.
When heat is transferred to the surrounding,$q = -Y \ J$.
Therefore,$\Delta U = (-Y) + (+X) = X - Y \ J$.
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201
ChemistryMediumMCQMHT CET · 2022
Which of the following pairs of elements have one electron in the $5d$-subshell in their observed electronic configuration?
A
$Sm (Z=62)$ and $Eu (Z=63)$
B
$Lu (Z=71)$ and $Dy (Z=66)$
C
$Gd (Z=64)$ and $Lu (Z=71)$
D
$Ce (Z=58)$ and $Nd (Z=60)$
Solution
(C) The electronic configurations of the given lanthanoids are as follows:
$Ce (Z=58) = [Xe] 4f^1 5d^1 6s^2$
$Nd (Z=60) = [Xe] 4f^4 6s^2$
$Sm (Z=62) = [Xe] 4f^6 6s^2$
$Eu (Z=63) = [Xe] 4f^7 6s^2$
$Gd (Z=64) = [Xe] 4f^7 5d^1 6s^2$
$Dy (Z=66) = [Xe] 4f^{10} 6s^2$
$Lu (Z=71) = [Xe] 4f^{14} 5d^1 6s^2$
Among the given options,$Gd (Z=64)$ and $Lu (Z=71)$ both possess exactly one electron in the $5d$-subshell.
$Ce (Z=58) = [Xe] 4f^1 5d^1 6s^2$
$Nd (Z=60) = [Xe] 4f^4 6s^2$
$Sm (Z=62) = [Xe] 4f^6 6s^2$
$Eu (Z=63) = [Xe] 4f^7 6s^2$
$Gd (Z=64) = [Xe] 4f^7 5d^1 6s^2$
$Dy (Z=66) = [Xe] 4f^{10} 6s^2$
$Lu (Z=71) = [Xe] 4f^{14} 5d^1 6s^2$
Among the given options,$Gd (Z=64)$ and $Lu (Z=71)$ both possess exactly one electron in the $5d$-subshell.
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202
ChemistryEasyMCQMHT CET · 2022
What is the number of unpaired electrons present in $f$-orbital at $+3$ oxidation state of $Lu (Z=71)$?
A
$0$
B
$4$
C
$2$
D
$7$
Solution
(A) The atomic number of Lutetium $(Lu)$ is $71$.
The electronic configuration of $Lu$ is $[Xe] 4f^{14} 5d^1 6s^2$.
In the $+3$ oxidation state,$Lu$ loses three electrons (two from $6s$ and one from $5d$).
The electronic configuration of $Lu^{3+}$ is $[Xe] 4f^{14}$.
Since the $4f$ subshell is completely filled with $14$ electrons,there are no unpaired electrons present.
Therefore,the number of unpaired electrons is $0$.
The electronic configuration of $Lu$ is $[Xe] 4f^{14} 5d^1 6s^2$.
In the $+3$ oxidation state,$Lu$ loses three electrons (two from $6s$ and one from $5d$).
The electronic configuration of $Lu^{3+}$ is $[Xe] 4f^{14}$.
Since the $4f$ subshell is completely filled with $14$ electrons,there are no unpaired electrons present.
Therefore,the number of unpaired electrons is $0$.
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203
ChemistryEasyMCQMHT CET · 2022
Identify the outer electronic configuration of the $first$ element of the $first$ transition series.
A
$[Ar] 3d^1 4s^1$
B
$[Kr] 4d^0 5s^2$
C
$[Kr] 4d^1 5s^2$
D
$[Ar] 3d^1 4s^2$
Solution
(D) The $first$ transition series corresponds to the $3d$ series,which starts with Scandium $(Sc)$.
Scandium has an atomic number of $21$.
The electronic configuration is determined by filling the $1s, 2s, 2p, 3s, 3p, 4s,$ and $3d$ orbitals.
$Sc (Z=21) = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1$.
In terms of the noble gas core,this is written as $[Ar] 3d^1 4s^2$.
Scandium has an atomic number of $21$.
The electronic configuration is determined by filling the $1s, 2s, 2p, 3s, 3p, 4s,$ and $3d$ orbitals.
$Sc (Z=21) = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1$.
In terms of the noble gas core,this is written as $[Ar] 3d^1 4s^2$.
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204
ChemistryMediumMCQMHT CET · 2022
Which of the following alloys is used for ultra-high speed flight?
A
Cupra-nickel
B
Bronze
C
Titanium alloy
D
Nichrome
Solution
(C) Titanium alloys are widely used in aerospace applications,particularly for ultra-high speed flight,because they maintain excellent mechanical properties and high strength-to-weight ratios at elevated temperatures.
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205
ChemistryEasyMCQMHT CET · 2022
Identify the element if its observed and expected electronic configuration is considered,specifically focusing on the transition metal series where $d$-orbital filling occurs.
A
$Ca$
B
$K$
C
$Sc$
D
$Rb$
Solution
(C) The element with atomic number $21$ is Scandium $(Sc)$.
Its electronic configuration is $[Ar] 3d^1 4s^2$.
In the transition metal series,the $3d$ orbital begins to fill after the $4s$ orbital is filled,which is a characteristic feature of transition elements.
Its electronic configuration is $[Ar] 3d^1 4s^2$.
In the transition metal series,the $3d$ orbital begins to fill after the $4s$ orbital is filled,which is a characteristic feature of transition elements.
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206
ChemistryEasyMCQMHT CET · 2022
Which element from the following exhibits a common oxidation state of $+2$ only?
A
$Cu$
B
$Co$
C
$Sc$
D
$Zn$
Solution
(D) The electronic configuration of $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$.
It loses two electrons from the $4s$ orbital to form the $Zn^{2+}$ ion,which has a stable fully-filled $3d^{10}$ configuration.
Therefore,$Zn$ exhibits only a $+2$ oxidation state.
It loses two electrons from the $4s$ orbital to form the $Zn^{2+}$ ion,which has a stable fully-filled $3d^{10}$ configuration.
Therefore,$Zn$ exhibits only a $+2$ oxidation state.
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207
ChemistryEasyMCQMHT CET · 2022
Which among the following elements in their respective oxidation states develops the lowest spin-only magnetic moment? (Atomic numbers: $Cu=29, Fe=26, Ni=28, Co=27$)
A
$Co^{2+}$
B
$Cu^{2+}$
C
$Fe^{2+}$
D
$Ni^{2+}$
Solution
(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1$. $Co^{2+}$ $(3d^7)$: $n=3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \text{ BM}$.
$2$. $Cu^{2+}$ $(3d^9)$: $n=1$,$\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73 \text{ BM}$.
$3$. $Fe^{2+}$ $(3d^6)$: $n=4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \text{ BM}$.
$4$. $Ni^{2+}$ $(3d^8)$: $n=2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \text{ BM}$.
Comparing the values,$Cu^{2+}$ has the lowest number of unpaired electrons $(n=1)$,hence it exhibits the lowest spin-only magnetic moment.
$1$. $Co^{2+}$ $(3d^7)$: $n=3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \text{ BM}$.
$2$. $Cu^{2+}$ $(3d^9)$: $n=1$,$\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73 \text{ BM}$.
$3$. $Fe^{2+}$ $(3d^6)$: $n=4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \text{ BM}$.
$4$. $Ni^{2+}$ $(3d^8)$: $n=2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \text{ BM}$.
Comparing the values,$Cu^{2+}$ has the lowest number of unpaired electrons $(n=1)$,hence it exhibits the lowest spin-only magnetic moment.
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208
ChemistryMediumMCQMHT CET · 2022
Which of the following alloys is used to construct marine condenser tubes of ships?
A
Titanium alloys
B
Nichrome
C
Bronze
D
Cupro-nickel
Solution
(D) Cupro-nickel alloys (typically containing $70-90\%$ copper and $10-30\%$ nickel) are widely used for marine condenser tubes because they possess excellent resistance to corrosion and biofouling in seawater environments.
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209
ChemistryEasyMCQMHT CET · 2022
What is the number of unpaired electrons present in $Sc^{3+}$ and $Zn^{2+}$ ions respectively?
A
$1, 0$
B
$3, 1$
C
$2, 2$
D
$0, 0$
Solution
(D) The atomic number of $Sc$ is $21$. The electronic configuration of $Sc$ is $[Ar] \ 3d^1 \ 4s^2$. Thus,$Sc^{3+}$ is $[Ar]$,which has $0$ unpaired electrons.
The atomic number of $Zn$ is $30$. The electronic configuration of $Zn$ is $[Ar] \ 3d^{10} \ 4s^2$. Thus,$Zn^{2+}$ is $[Ar] \ 3d^{10}$,which has $0$ unpaired electrons.
Therefore,the number of unpaired electrons in $Sc^{3+}$ and $Zn^{2+}$ are $0$ and $0$ respectively.
The atomic number of $Zn$ is $30$. The electronic configuration of $Zn$ is $[Ar] \ 3d^{10} \ 4s^2$. Thus,$Zn^{2+}$ is $[Ar] \ 3d^{10}$,which has $0$ unpaired electrons.
Therefore,the number of unpaired electrons in $Sc^{3+}$ and $Zn^{2+}$ are $0$ and $0$ respectively.
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210
ChemistryEasyMCQMHT CET · 2022
Which among the following is an actinoid element?
A
$Gd$
B
$Lu$
C
$Pr$
D
$Pa$
Solution
(D) The actinoids are the elements with atomic numbers $89$ to $103$.
$Gd$ (Gadolinium,$Z=64$),$Lu$ (Lutetium,$Z=71$),and $Pr$ (Praseodymium,$Z=59$) are lanthanoids.
$Pa$ (Protactinium,$Z=91$) is an actinoid element.
$Gd$ (Gadolinium,$Z=64$),$Lu$ (Lutetium,$Z=71$),and $Pr$ (Praseodymium,$Z=59$) are lanthanoids.
$Pa$ (Protactinium,$Z=91$) is an actinoid element.
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211
ChemistryEasyMCQMHT CET · 2022
Which among the following lanthanoids forms the most stable $+2$ oxidation state?
A
$Eu$ $(Z=63)$
B
$Ce$ $(Z=58)$
C
$Pm$ $(Z=61)$
D
$Gd$ $(Z=64)$
Solution
(A) The electronic configuration of $Eu$ $(Z=63)$ is $[Xe] 4f^7 6s^2$.
When $Eu$ loses two electrons to form $Eu^{2+}$,its configuration becomes $[Xe] 4f^7$.
This $4f^7$ configuration is half-filled,which provides extra stability to the $Eu^{2+}$ ion.
When $Eu$ loses two electrons to form $Eu^{2+}$,its configuration becomes $[Xe] 4f^7$.
This $4f^7$ configuration is half-filled,which provides extra stability to the $Eu^{2+}$ ion.
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212
ChemistryEasyMCQMHT CET · 2022
Identify the strongest base from the following.
A
$Sm(OH)_3$
B
$La(OH)_3$
C
$Lu(OH)_3$
D
$Pr(OH)_3$
Solution
(B) In the case of lanthanides,as the atomic number $(Z)$ increases,the ionic size decreases due to lanthanide contraction.
As the size of the $Ln^{3+}$ ion decreases,the covalent character of the $Ln-OH$ bond increases,which leads to a decrease in basic strength.
Therefore,the basic character follows the order: $La(OH)_3 > Pr(OH)_3 > Sm(OH)_3 > Lu(OH)_3$.
Thus,$La(OH)_3$ is the strongest base.
As the size of the $Ln^{3+}$ ion decreases,the covalent character of the $Ln-OH$ bond increases,which leads to a decrease in basic strength.
Therefore,the basic character follows the order: $La(OH)_3 > Pr(OH)_3 > Sm(OH)_3 > Lu(OH)_3$.
Thus,$La(OH)_3$ is the strongest base.
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213
ChemistryMediumMCQMHT CET · 2022
Which among the following elements does not have any electron in the $f-$subshell in either its expected or observed electronic configuration?
A
$Lu (Z=71)$
B
$La (Z=57)$
C
$Eu (Z=63)$
D
$Nd (Z=60)$
Solution
(B) The electronic configuration of $La (Z=57)$ is $[Xe] 5d^1 6s^2$.
It does not contain any electrons in the $4f$ subshell.
For the other elements:
$Nd (Z=60)$ is $[Xe] 4f^3 6s^2$.
$Eu (Z=63)$ is $[Xe] 4f^7 6s^2$.
$Lu (Z=71)$ is $[Xe] 4f^{14} 5d^1 6s^2$.
Thus,$La$ is the correct answer.
It does not contain any electrons in the $4f$ subshell.
For the other elements:
$Nd (Z=60)$ is $[Xe] 4f^3 6s^2$.
$Eu (Z=63)$ is $[Xe] 4f^7 6s^2$.
$Lu (Z=71)$ is $[Xe] 4f^{14} 5d^1 6s^2$.
Thus,$La$ is the correct answer.
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214
ChemistryEasyMCQMHT CET · 2022
Identify the element with the smallest ionic radius in the $+3$ oxidation state from the following:
A
$Yb^{3+}$
B
$Eu^{3+}$
C
$Er^{3+}$
D
$Lu^{3+}$
Solution
(D) In the lanthanoid series,as the atomic number $(Z)$ increases,the ionic radius decreases due to lanthanoid contraction.
Since $Lu$ $(Z = 71)$ has the highest atomic number among the given elements $(Eu: 63, Er: 68, Yb: 70, Lu: 71)$,it has the smallest ionic radius in the $+3$ oxidation state.
Since $Lu$ $(Z = 71)$ has the highest atomic number among the given elements $(Eu: 63, Er: 68, Yb: 70, Lu: 71)$,it has the smallest ionic radius in the $+3$ oxidation state.
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215
ChemistryMediumMCQMHT CET · 2022
Which among the following properties of lanthanoids is $NOT$ true?
A
Have greater coordination number than six
B
All are non-radioactive elements
C
Good conductors of heat and electricity
D
Strongly paramagnetic
Solution
(B) Lanthanoids are generally non-radioactive,except for $Pm$ (Promethium),which is radioactive. Therefore,the statement that all lanthanoids are non-radioactive is incorrect.
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216
ChemistryMediumMCQMHT CET · 2022
Which element from the following has a half-filled $f$ orbital in the $+3$ oxidation state?
A
$U (Z=92)$
B
$Lr (Z=103)$
C
$Cm (Z=96)$
D
$Th (Z=90)$
Solution
(C) The electronic configuration of $Cm (Z=96)$ is $[Rn] 5f^7 6d^1 7s^2$.
In the $+3$ oxidation state,it loses three electrons to form $Cm^{3+}$,which has the configuration $[Rn] 5f^7$.
Since the $f$ orbital can hold a maximum of $14$ electrons,$f^7$ represents a half-filled $f$ orbital.
In the $+3$ oxidation state,it loses three electrons to form $Cm^{3+}$,which has the configuration $[Rn] 5f^7$.
Since the $f$ orbital can hold a maximum of $14$ electrons,$f^7$ represents a half-filled $f$ orbital.
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217
ChemistryEasyMCQMHT CET · 2022
Identify the non-radioactive element from the following.
A
$Nd$
B
$Np$
C
$U$
D
$Cf$
Solution
(A) The elements $Np$ (Neptunium),$U$ (Uranium),and $Cf$ (Californium) are all radioactive actinoids.
$Nd$ (Neodymium) is a lanthanoid element and is non-radioactive (stable).
$Nd$ (Neodymium) is a lanthanoid element and is non-radioactive (stable).
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218
ChemistryMediumMCQMHT CET · 2022
Which of the following equations is correct for the relation between standard cell potential and equilibrium constant?
A
$E_{cell}^0 = \log_{10} K \frac{n}{0.0592}$
B
$E_{cell}^0 = \frac{0.0592}{n} \log_{10} K$
C
$E_{cell} = \frac{0.0592}{n} \log_{10} K$
D
$E_{cell} = \log_{10} K \frac{n}{0.0592}$
Solution
(B) We know the Nernst equation: $E_{cell} = E_{cell}^0 - \frac{0.0592}{n} \log_{10} Q$.
At equilibrium,$E_{cell} = 0$ and $Q = K$.
Substituting these values,we get: $0 = E_{cell}^0 - \frac{0.0592}{n} \log_{10} K$.
Therefore,the correct relation is: $E_{cell}^0 = \frac{0.0592}{n} \log_{10} K$.
At equilibrium,$E_{cell} = 0$ and $Q = K$.
Substituting these values,we get: $0 = E_{cell}^0 - \frac{0.0592}{n} \log_{10} K$.
Therefore,the correct relation is: $E_{cell}^0 = \frac{0.0592}{n} \log_{10} K$.
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219
ChemistryMediumMCQMHT CET · 2022
What is the standard reduction potential of $Cu^{2+} \mid Cu_{(s)}$ if the $E^{\circ}$ of the following cell is $0.46 \ V$ (in $V$)?
$Cu_{(s)} \mid Cu_{(aq)}^{2+} \parallel Ag_{(aq)}^{+} \mid Ag_{(s)}$ ; $E_{Ag^{+} / Ag}^{\circ} = 0.80 \ V$
$Cu_{(s)} \mid Cu_{(aq)}^{2+} \parallel Ag_{(aq)}^{+} \mid Ag_{(s)}$ ; $E_{Ag^{+} / Ag}^{\circ} = 0.80 \ V$
A
$1.56$
B
$0.34$
C
$1.44$
D
$1.26$
Solution
(B) The standard cell potential is given by the formula: $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$.
In the given cell,$Ag$ is the cathode and $Cu$ is the anode.
So,$E_{\text{cell}}^{\circ} = E_{Ag^{+}/Ag}^{\circ} - E_{Cu^{2+}/Cu}^{\circ}$.
Substituting the given values: $0.46 \ V = 0.80 \ V - E_{Cu^{2+}/Cu}^{\circ}$.
Rearranging the equation: $E_{Cu^{2+}/Cu}^{\circ} = 0.80 \ V - 0.46 \ V = 0.34 \ V$.
In the given cell,$Ag$ is the cathode and $Cu$ is the anode.
So,$E_{\text{cell}}^{\circ} = E_{Ag^{+}/Ag}^{\circ} - E_{Cu^{2+}/Cu}^{\circ}$.
Substituting the given values: $0.46 \ V = 0.80 \ V - E_{Cu^{2+}/Cu}^{\circ}$.
Rearranging the equation: $E_{Cu^{2+}/Cu}^{\circ} = 0.80 \ V - 0.46 \ V = 0.34 \ V$.
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220
ChemistryMediumMCQMHT CET · 2022
Calculate the quantity of electricity required to produce $0.42 \ g$ of $Ag$ at the cathode during the electrolysis of an $AgNO_3$ solution. (Molar mass of $Ag = 108 \ g \ mol^{-1}$) (in $C$)
A
$965.0$
B
$470.0$
C
$257.1$
D
$375.3$
Solution
(D) The reduction reaction at the cathode is: $Ag^+ + e^- \rightarrow Ag(s)$.
According to Faraday's law of electrolysis,the mass deposited $(W)$ is given by $W = \frac{M \times Q}{n \times F}$,where $M$ is the molar mass,$Q$ is the charge,$n$ is the number of electrons involved ($n=1$ for $Ag^+$),and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.
Substituting the given values: $0.42 = \frac{108 \times Q}{1 \times 96500}$.
Solving for $Q$: $Q = \frac{0.42 \times 96500}{108}$.
$Q = 375.277 \ C \approx 375.3 \ C$.
According to Faraday's law of electrolysis,the mass deposited $(W)$ is given by $W = \frac{M \times Q}{n \times F}$,where $M$ is the molar mass,$Q$ is the charge,$n$ is the number of electrons involved ($n=1$ for $Ag^+$),and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.
Substituting the given values: $0.42 = \frac{108 \times Q}{1 \times 96500}$.
Solving for $Q$: $Q = \frac{0.42 \times 96500}{108}$.
$Q = 375.277 \ C \approx 375.3 \ C$.
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221
ChemistryEasyMCQMHT CET · 2022
Calculate the standard cell potential for a cell having the following reaction: $2 Al_{(s)} + 3 Ni^{2+} \rightarrow 2 Al^{3+} + 3 Ni_{(s)}$ given that $E_{Ni^{2+}/Ni}^{\circ} = -0.25 \ V$ and $E_{Al^{3+}/Al}^{\circ} = -1.66 \ V$. (in $V$)
A
$0.50$
B
$1.41$
C
$-0.50$
D
$0.41$
Solution
(B) The standard cell potential is calculated using the formula: $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$.
In the given reaction,$Al$ is oxidized (anode) and $Ni^{2+}$ is reduced (cathode).
$E_{\text{cathode}}^{\circ} = E_{Ni^{2+}/Ni}^{\circ} = -0.25 \ V$.
$E_{\text{anode}}^{\circ} = E_{Al^{3+}/Al}^{\circ} = -1.66 \ V$.
$E_{\text{cell}}^{\circ} = -0.25 \ V - (-1.66 \ V) = -0.25 + 1.66 = 1.41 \ V$.
In the given reaction,$Al$ is oxidized (anode) and $Ni^{2+}$ is reduced (cathode).
$E_{\text{cathode}}^{\circ} = E_{Ni^{2+}/Ni}^{\circ} = -0.25 \ V$.
$E_{\text{anode}}^{\circ} = E_{Al^{3+}/Al}^{\circ} = -1.66 \ V$.
$E_{\text{cell}}^{\circ} = -0.25 \ V - (-1.66 \ V) = -0.25 + 1.66 = 1.41 \ V$.
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222
ChemistryMediumMCQMHT CET · 2022
Which of the following cannot be considered as an application of the electrochemical series?
A
Spontaneity of redox reaction
B
Relative strength of reducing agent
C
Relative strength of oxidizing agent
D
Use of $SHE$ in automobile battery
Solution
(D) The electrochemical series is used to predict the spontaneity of redox reactions,the relative strength of oxidizing agents,and the relative strength of reducing agents.
However,the use of $SHE$ (Standard Hydrogen Electrode) in an automobile battery is not an application of the electrochemical series. $SHE$ is a reference electrode used to measure standard electrode potentials,whereas automobile batteries (like lead-acid batteries) rely on specific chemical reactions to store and release electrical energy.
However,the use of $SHE$ (Standard Hydrogen Electrode) in an automobile battery is not an application of the electrochemical series. $SHE$ is a reference electrode used to measure standard electrode potentials,whereas automobile batteries (like lead-acid batteries) rely on specific chemical reactions to store and release electrical energy.
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223
ChemistryMediumMCQMHT CET · 2022
What is the change in potential of the following cell $Zn_{(s)}|Zn^{2+} (1 \ M)||Pb^{2+} (1 \ M)|Pb_{(s)}$ if the concentration of ions at the anode is increased $10$ times?
A
Decreases by $0.0296 \ V$
B
Increases by $10 \ V$
C
Decreases by $10 \ V$
D
Increases by $0.0296 \ V$
Solution
(A) The cell reaction is $Zn_{(s)} + Pb^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Pb_{(s)}$.
Using the Nernst equation: $E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Pb^{2+}]}$.
Here,$n = 2$. Initially,$[Zn^{2+}] = 1 \ M$ and $[Pb^{2+}] = 1 \ M$,so $E_{\text{cell, initial}} = E_{\text{cell}}^0$.
When the concentration of $Zn^{2+}$ at the anode increases $10$ times,$[Zn^{2+}] = 10 \ M$.
The new potential is $E_{\text{cell, final}} = E_{\text{cell}}^0 - \frac{0.0591}{2} \log \frac{10}{1} = E_{\text{cell}}^0 - 0.02955 \log(10) = E_{\text{cell}}^0 - 0.0296 \ V$.
Therefore,the potential decreases by $0.0296 \ V$.
Using the Nernst equation: $E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Pb^{2+}]}$.
Here,$n = 2$. Initially,$[Zn^{2+}] = 1 \ M$ and $[Pb^{2+}] = 1 \ M$,so $E_{\text{cell, initial}} = E_{\text{cell}}^0$.
When the concentration of $Zn^{2+}$ at the anode increases $10$ times,$[Zn^{2+}] = 10 \ M$.
The new potential is $E_{\text{cell, final}} = E_{\text{cell}}^0 - \frac{0.0591}{2} \log \frac{10}{1} = E_{\text{cell}}^0 - 0.02955 \log(10) = E_{\text{cell}}^0 - 0.0296 \ V$.
Therefore,the potential decreases by $0.0296 \ V$.
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224
ChemistryMediumMCQMHT CET · 2022
How many coulombs of electricity is required to produce $1 \ g$ of $Na$ metal from its ions (in $C$)? (Atomic mass of $Na = 23 \ g/mol$)
A
$1516$
B
$3132$
C
$2088$
D
$4196$
Solution
(D) The reduction reaction is: $Na^{+} + e^{-} \rightarrow Na_{(s)}$
From the stoichiometry,$1 \ mol$ of $Na$ $(23 \ g)$ requires $1 \ mol$ of electrons ($96500 \ C$ of charge).
Therefore,the charge required to produce $1 \ g$ of $Na$ is:
$Q = \frac{1 \ g}{23 \ g/mol} \times 96500 \ C/mol$
$Q = 4195.65 \ C \approx 4196 \ C$
From the stoichiometry,$1 \ mol$ of $Na$ $(23 \ g)$ requires $1 \ mol$ of electrons ($96500 \ C$ of charge).
Therefore,the charge required to produce $1 \ g$ of $Na$ is:
$Q = \frac{1 \ g}{23 \ g/mol} \times 96500 \ C/mol$
$Q = 4195.65 \ C \approx 4196 \ C$
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225
ChemistryMediumMCQMHT CET · 2022
Calculate the mass of a divalent metal produced at the cathode by passing $5 \ A$ current through its salt solution for $100 \ minutes$. (Molar mass of metal $= x \ g/mol$)
A
$\frac{193}{30} x \ g$
B
$\frac{193}{15} x \ g$
C
$\frac{15}{193} x \ g$
D
$\frac{30}{193} x \ g$
Solution
(D) According to Faraday's law of electrolysis,the mass $W$ deposited is given by $W = \frac{M \times I \times t}{n \times F}$.
Here,$M = x \ g/mol$,$I = 5 \ A$,$t = 100 \times 60 \ s = 6000 \ s$,$n = 2$ (for a divalent metal),and $F \approx 96500 \ C/mol$.
Substituting the values: $W = \frac{x \times 5 \times 6000}{2 \times 96500}$.
$W = \frac{30000 x}{193000} = \frac{30}{193} x \ g$.
Here,$M = x \ g/mol$,$I = 5 \ A$,$t = 100 \times 60 \ s = 6000 \ s$,$n = 2$ (for a divalent metal),and $F \approx 96500 \ C/mol$.
Substituting the values: $W = \frac{x \times 5 \times 6000}{2 \times 96500}$.
$W = \frac{30000 x}{193000} = \frac{30}{193} x \ g$.
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226
ChemistryMediumMCQMHT CET · 2022
Identify the reaction taking place at the anode during the electrolysis of molten $NaCl$.
A
$2 OH^{-} \rightarrow H_2O + \frac{1}{2} O_2 + 2 e^{-}$
B
$2 H^{+} + 2 e^{-} \rightarrow H_2$
C
$Na^{+} + 1 e^{-} \rightarrow Na_{(s)}$
D
$2 Cl^{-} \rightarrow Cl_{2(g)} + 2 e^{-}$
Solution
(D) During the electrolysis of molten $NaCl$,the ions present are $Na^{+}$ and $Cl^{-}$.
At the anode,oxidation occurs,which involves the loss of electrons.
The $Cl^{-}$ ions migrate to the anode and undergo oxidation to form chlorine gas:
$2 Cl^{-} \rightarrow Cl_{2(g)} + 2 e^{-}$
At the anode,oxidation occurs,which involves the loss of electrons.
The $Cl^{-}$ ions migrate to the anode and undergo oxidation to form chlorine gas:
$2 Cl^{-} \rightarrow Cl_{2(g)} + 2 e^{-}$
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227
ChemistryMediumMCQMHT CET · 2022
What will be the weight of $Al$ deposited at the cathode when $0.5 \ F$ of electricity is passed through an aqueous solution of $AlCl_3$ (in $g$)? (Atomic mass of $Al = 27$)
A
$4.5$
B
$13.5$
C
$40.5$
D
$3.0$
Solution
(A) The reduction reaction at the cathode is: $Al^{3+} + 3e^- \rightarrow Al$.
From the reaction,$3 \ F$ of electricity deposits $1 \ \text{mole}$ of $Al$ $(27 \ g)$.
Therefore,$0.5 \ F$ of electricity will deposit:
$\text{Weight} = \frac{\text{Atomic mass} \times \text{Faradays}}{n} = \frac{27 \times 0.5}{3} = 4.5 \ g$.
From the reaction,$3 \ F$ of electricity deposits $1 \ \text{mole}$ of $Al$ $(27 \ g)$.
Therefore,$0.5 \ F$ of electricity will deposit:
$\text{Weight} = \frac{\text{Atomic mass} \times \text{Faradays}}{n} = \frac{27 \times 0.5}{3} = 4.5 \ g$.
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228
ChemistryEasyMCQMHT CET · 2022
In a certain electrolysis experiment,$0.650 \ g$ of zinc is deposited in a cell having $ZnSO_4$ solution. Calculate the mass of $Cu$ deposited in another cell having $CuSO_4$ solution arranged in series with the first cell (in $g$)? (Molar mass of $Zn = 65 \ g \ mol^{-1}$,$Cu = 63.5 \ g \ mol^{-1}$)
A
$0.635$
B
$6.35$
C
$0.317$
D
$3.17$
Solution
(A) According to Faraday's second law of electrolysis,when cells are connected in series,the mass of substances deposited is proportional to their equivalent masses: $\frac{W_{Zn}}{E_{Zn}} = \frac{W_{Cu}}{E_{Cu}}$
Equivalent mass of $Zn = \frac{\text{Molar mass}}{\text{Valency factor}} = \frac{65}{2} = 32.5 \ g \ eq^{-1}$
Equivalent mass of $Cu = \frac{\text{Molar mass}}{\text{Valency factor}} = \frac{63.5}{2} = 31.75 \ g \ eq^{-1}$
Substituting the values: $\frac{0.650}{32.5} = \frac{W_{Cu}}{31.75}$
$W_{Cu} = \frac{0.650 \times 31.75}{32.5} = 0.635 \ g$
Equivalent mass of $Zn = \frac{\text{Molar mass}}{\text{Valency factor}} = \frac{65}{2} = 32.5 \ g \ eq^{-1}$
Equivalent mass of $Cu = \frac{\text{Molar mass}}{\text{Valency factor}} = \frac{63.5}{2} = 31.75 \ g \ eq^{-1}$
Substituting the values: $\frac{0.650}{32.5} = \frac{W_{Cu}}{31.75}$
$W_{Cu} = \frac{0.650 \times 31.75}{32.5} = 0.635 \ g$
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229
ChemistryEasyMCQMHT CET · 2022
Identify the representation of the standard hydrogen electrode.
A
$H^{+}_{(aq)} | H_2(1 \ M) | Pt, [H^{+}]$
B
$Pt | H_2(g, 1 \ atm) | H^{+}(1 \ M)$
C
$H_{2(g)} | H^{+}_{(aq)} | Pt, H^{+}$
D
$Pt, [H^{+}] H_{2(g)} | H^{+} | H^{+}(1 \ M)$
Solution
(B) The standard hydrogen electrode $(SHE)$ is represented as $Pt | H_2(g, 1 \ atm) | H^{+}(1 \ M)$ when it acts as an anode or $H^{+}(1 \ M) | H_2(g, 1 \ atm) | Pt$ when it acts as a cathode.
Given the options,the standard representation for the electrode is $Pt | H_2(g, 1 \ atm) | H^{+}(1 \ M)$.
Under standard conditions,the concentration of $H^{+}$ ions is $1 \ M$ and the pressure of $H_2$ gas is $1 \ atm$ (or $1 \ bar$).
Given the options,the standard representation for the electrode is $Pt | H_2(g, 1 \ atm) | H^{+}(1 \ M)$.
Under standard conditions,the concentration of $H^{+}$ ions is $1 \ M$ and the pressure of $H_2$ gas is $1 \ atm$ (or $1 \ bar$).
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230
ChemistryEasyMCQMHT CET · 2022
How many Faraday of electricity is required to produce $5 \ g$ of magnesium from magnesium chloride (in $F$)? (Molar mass $Mg = 24 \ g \ mol^{-1}$)
A
$2.451$
B
$0.417$
C
$6.0$
D
$9.634$
Solution
(B) The chemical reaction for the reduction of magnesium ions is: $Mg^{2+} + 2e^{-} \rightarrow Mg$
From the stoichiometry,$1 \ mol$ of $Mg$ requires $2 \ F$ of electricity.
The number of moles of $Mg$ produced is: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \ g}{24 \ g \ mol^{-1}} = 0.2083 \ mol$.
Therefore,the Faraday of electricity required is: $0.2083 \ mol \times 2 \ F \ mol^{-1} = 0.4166 \ F \approx 0.417 \ F$.
From the stoichiometry,$1 \ mol$ of $Mg$ requires $2 \ F$ of electricity.
The number of moles of $Mg$ produced is: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \ g}{24 \ g \ mol^{-1}} = 0.2083 \ mol$.
Therefore,the Faraday of electricity required is: $0.2083 \ mol \times 2 \ F \ mol^{-1} = 0.4166 \ F \approx 0.417 \ F$.
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231
ChemistryEasyMCQMHT CET · 2022
How long will it take to produce $5.4 \ g$ of $Ag$ from molten $AgCl$ by passing $5 \ A$ current (in $s$)? (Molar mass $Ag = 108 \ g \ mol^{-1}$)
A
$1930$
B
$965$
C
$193$
D
$9650$
Solution
(B) The reaction for the deposition of $Ag$ is: $Ag^+ + e^- \rightarrow Ag(s)$.
Here,$n = 1$ (number of electrons involved).
Using Faraday's law of electrolysis: $W = \frac{M \times I \times t}{n \times F}$.
Given: $W = 5.4 \ g$,$M = 108 \ g \ mol^{-1}$,$I = 5 \ A$,$F = 96500 \ C \ mol^{-1}$.
Substituting the values: $5.4 = \frac{108 \times 5 \times t}{1 \times 96500}$.
$t = \frac{5.4 \times 96500}{108 \times 5}$.
$t = \frac{521100}{540} = 965 \ s$.
Here,$n = 1$ (number of electrons involved).
Using Faraday's law of electrolysis: $W = \frac{M \times I \times t}{n \times F}$.
Given: $W = 5.4 \ g$,$M = 108 \ g \ mol^{-1}$,$I = 5 \ A$,$F = 96500 \ C \ mol^{-1}$.
Substituting the values: $5.4 = \frac{108 \times 5 \times t}{1 \times 96500}$.
$t = \frac{5.4 \times 96500}{108 \times 5}$.
$t = \frac{521100}{540} = 965 \ s$.
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232
ChemistryEasyMCQMHT CET · 2022
What current strength is required to deposit $36 \ g$ of $Ag$ in $7 \ minute$ from $AgNO_3$ solution by electrolysis (in $A$)? (Atomic mass $Ag = 108$)
A
$11.44$
B
$5.72$
C
$76.6$
D
$38.3$
Solution
(C) According to Faraday's law of electrolysis,$W = \frac{M \times I \times t}{n \times F}$.
Here,$W = 36 \ g$,$M = 108 \ g/mol$,$n = 1$ (for $Ag^+ + e^- \rightarrow Ag$),$t = 7 \times 60 \ s = 420 \ s$,and $F = 96500 \ C/mol$.
Substituting the values: $36 = \frac{108 \times I \times 420}{1 \times 96500}$.
$I = \frac{36 \times 96500}{108 \times 420}$.
$I = \frac{3474000}{45360} \approx 76.58 \ A \approx 76.6 \ A$.
Here,$W = 36 \ g$,$M = 108 \ g/mol$,$n = 1$ (for $Ag^+ + e^- \rightarrow Ag$),$t = 7 \times 60 \ s = 420 \ s$,and $F = 96500 \ C/mol$.
Substituting the values: $36 = \frac{108 \times I \times 420}{1 \times 96500}$.
$I = \frac{36 \times 96500}{108 \times 420}$.
$I = \frac{3474000}{45360} \approx 76.58 \ A \approx 76.6 \ A$.
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233
ChemistryEasyMCQMHT CET · 2022
Calculate the standard potential of a cell having the following electrode reactions:
$Cd_{(aq)}^{2+} + 2e^{-} \rightarrow Cd_{(s)}$ $E^{\circ} = -0.403 \ V$
$Zn_{(aq)}^{2+} + 2e^{-} \rightarrow Zn_{(s)}$ $E^{\circ} = -0.763 \ V$ (in $V$)
$Cd_{(aq)}^{2+} + 2e^{-} \rightarrow Cd_{(s)}$ $E^{\circ} = -0.403 \ V$
$Zn_{(aq)}^{2+} + 2e^{-} \rightarrow Zn_{(s)}$ $E^{\circ} = -0.763 \ V$ (in $V$)
A
$0.201$
B
$0.360$
C
$0.481$
D
$1.166$
Solution
(B) The standard cell potential is calculated using the formula: $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$.
In this cell,$Cd^{2+}/Cd$ acts as the cathode (higher reduction potential) and $Zn^{2+}/Zn$ acts as the anode (lower reduction potential).
Substituting the values: $E_{cell}^{\circ} = -0.403 \ V - (-0.763 \ V)$.
$E_{cell}^{\circ} = -0.403 + 0.763 = 0.360 \ V$.
In this cell,$Cd^{2+}/Cd$ acts as the cathode (higher reduction potential) and $Zn^{2+}/Zn$ acts as the anode (lower reduction potential).
Substituting the values: $E_{cell}^{\circ} = -0.403 \ V - (-0.763 \ V)$.
$E_{cell}^{\circ} = -0.403 + 0.763 = 0.360 \ V$.
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234
ChemistryEasyMCQMHT CET · 2022
Which of the following is $NOT$ a function of a salt bridge?
A
Provide electrical contact
B
Convert electrical energy to chemical energy
C
Maintain electrical neutrality
D
Prevent mixing of solutions
Solution
(B) salt bridge is a device used in an electrochemical cell to connect its oxidation and reduction half-cells. Its primary functions are:
$1$. To complete the electrical circuit by providing electrical contact between the two solutions.
$2$. To maintain electrical neutrality in both half-cells by allowing the migration of ions.
$3$. To prevent the direct mixing of the two electrolyte solutions.
It does not convert electrical energy into chemical energy; rather,it facilitates the flow of current in a galvanic cell where chemical energy is converted into electrical energy.
$1$. To complete the electrical circuit by providing electrical contact between the two solutions.
$2$. To maintain electrical neutrality in both half-cells by allowing the migration of ions.
$3$. To prevent the direct mixing of the two electrolyte solutions.
It does not convert electrical energy into chemical energy; rather,it facilitates the flow of current in a galvanic cell where chemical energy is converted into electrical energy.
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235
ChemistryMediumMCQMHT CET · 2022
Which of the following is $NOT$ an example of a secondary voltaic cell?
A
Nickel-cadmium cell
B
Dry cell
C
Lead storage battery
D
Mercury cell
Solution
(B) secondary voltaic cell is a rechargeable battery.
$A$,$C$,and $D$ are examples of secondary cells or rechargeable systems.
$B$ (Dry cell) is a primary cell because it cannot be recharged.
$A$,$C$,and $D$ are examples of secondary cells or rechargeable systems.
$B$ (Dry cell) is a primary cell because it cannot be recharged.
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236
ChemistryEasyMCQMHT CET · 2022
What is the $SI$ unit of molar conductivity?
A
$S \ dm^3 \ mol^{-1}$
B
$S \ m^2 \ mol^{-1}$
C
$S \ cm^2 \ mol^{-1}$
D
$S \ m^2$
Solution
(B) Molar conductivity $(\Lambda_m)$ is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in a solution.
The formula is $\Lambda_m = \frac{\kappa}{c}$,where $\kappa$ is conductivity $(S \ m^{-1})$ and $c$ is concentration $(mol \ m^{-3})$.
Substituting the units: $\frac{S \ m^{-1}}{mol \ m^{-3}} = S \ m^2 \ mol^{-1}$.
Thus,the $SI$ unit of molar conductivity is $S \ m^2 \ mol^{-1}$.
The formula is $\Lambda_m = \frac{\kappa}{c}$,where $\kappa$ is conductivity $(S \ m^{-1})$ and $c$ is concentration $(mol \ m^{-3})$.
Substituting the units: $\frac{S \ m^{-1}}{mol \ m^{-3}} = S \ m^2 \ mol^{-1}$.
Thus,the $SI$ unit of molar conductivity is $S \ m^2 \ mol^{-1}$.
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237
ChemistryMediumMCQMHT CET · 2022
Which among the following concentrations of $KCl$ solution is not used to determine the cell constant of a conductivity cell?
A
Saturated $KCl$
B
$1.0 \ M \ KCl$
C
$0.01 \ M \ KCl$
D
$0.1 \ M \ KCl$
Solution
(A) The cell constant $(G^*)$ of a conductivity cell is determined by measuring the resistance of a standard $KCl$ solution whose conductivity $(\kappa)$ is already known.
Standard solutions typically used for this purpose are $0.01 \ M$,$0.1 \ M$,and $1.0 \ M$ $KCl$ solutions.
$A$ saturated $KCl$ solution is not used because its concentration is extremely high,leading to very high conductivity and potential errors in resistance measurement due to polarization effects and instrument limitations.
Standard solutions typically used for this purpose are $0.01 \ M$,$0.1 \ M$,and $1.0 \ M$ $KCl$ solutions.
$A$ saturated $KCl$ solution is not used because its concentration is extremely high,leading to very high conductivity and potential errors in resistance measurement due to polarization effects and instrument limitations.
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238
ChemistryMediumMCQMHT CET · 2022
The conductivity of $0.20 \ M \ KCl$ solution at $300 \ K$ is $0.0248 \ \Omega^{-1} \ cm^{-1}$. What is its molar conductivity?
A
$124 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
B
$93 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
C
$62 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
D
$186 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Solution
(A) The formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
Given:
Conductivity $\kappa = 0.0248 \ \Omega^{-1} \ cm^{-1}$
Molarity $M = 0.20 \ M$
Substituting the values:
$\Lambda_{m} = \frac{0.0248 \times 1000}{0.20}$
$\Lambda_{m} = \frac{24.8}{0.20} = 124 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Given:
Conductivity $\kappa = 0.0248 \ \Omega^{-1} \ cm^{-1}$
Molarity $M = 0.20 \ M$
Substituting the values:
$\Lambda_{m} = \frac{0.0248 \times 1000}{0.20}$
$\Lambda_{m} = \frac{24.8}{0.20} = 124 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
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239
ChemistryDifficultMCQMHT CET · 2022
Calculate molar conductivity at infinite dilution for $NaBr$ if molar conductivity at infinite dilution for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.
A
$128 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
B
$302 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
C
$278 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
D
$176 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Solution
(A) According to Kohlrausch's law of independent migration of ions:
$\wedge^0_{NaBr} = \lambda^0_{Na^+} + \lambda^0_{Br^-}$
$\wedge^0_{NaCl} = \lambda^0_{Na^+} + \lambda^0_{Cl^-} = 126 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\wedge^0_{KBr} = \lambda^0_{K^+} + \lambda^0_{Br^-} = 152 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\wedge^0_{KCl} = \lambda^0_{K^+} + \lambda^0_{Cl^-} = 150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
To obtain $\wedge^0_{NaBr}$,we perform the operation:
$\wedge^0_{NaBr} = \wedge^0_{NaCl} + \wedge^0_{KBr} - \wedge^0_{KCl}$
$\wedge^0_{NaBr} = 126 + 152 - 150 = 128 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\wedge^0_{NaBr} = \lambda^0_{Na^+} + \lambda^0_{Br^-}$
$\wedge^0_{NaCl} = \lambda^0_{Na^+} + \lambda^0_{Cl^-} = 126 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\wedge^0_{KBr} = \lambda^0_{K^+} + \lambda^0_{Br^-} = 152 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\wedge^0_{KCl} = \lambda^0_{K^+} + \lambda^0_{Cl^-} = 150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
To obtain $\wedge^0_{NaBr}$,we perform the operation:
$\wedge^0_{NaBr} = \wedge^0_{NaCl} + \wedge^0_{KBr} - \wedge^0_{KCl}$
$\wedge^0_{NaBr} = 126 + 152 - 150 = 128 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
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240
ChemistryMediumMCQMHT CET · 2022
For which of the following electrolytes is Kohlrausch's law of independent migration of ions used to calculate molar conductivity at zero concentration?
A
$NaNO_3$
B
$NH_4OH$
C
$KCl$
D
$BaSO_4$
Solution
(B) Kohlrausch's law of independent migration of ions is primarily used to determine the molar conductivity at infinite dilution (zero concentration) for weak electrolytes,such as $NH_4OH$,which do not dissociate completely in solution. For strong electrolytes like $NaNO_3$,$KCl$,and $BaSO_4$,the molar conductivity at zero concentration can be determined by extrapolating the plot of $\Lambda_m$ versus $\sqrt{c}$.
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241
ChemistryMediumMCQMHT CET · 2022
The resistance of a conductivity cell containing $0.001 \ M$ $KCl$ solution at $300 \ K$ is $150 \ \Omega$. What is the cell constant if conductivity of $KCl$ solution is $1.5 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$ (in $cm^{-1}$)?
A
$0.015$
B
$0.0225$
C
$0.0337$
D
$0.0450$
Solution
(B) The relationship between conductivity $(\kappa)$,resistance $(R)$,and cell constant $(G^* = \frac{L}{A})$ is given by: $\kappa = \frac{1}{R} \times G^*$.
Given: $\kappa = 1.5 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$ and $R = 150 \ \Omega$.
Substituting the values: $1.5 \times 10^{-4} = \frac{1}{150} \times G^*$.
Therefore,$G^* = 1.5 \times 10^{-4} \times 150 = 0.0225 \ cm^{-1}$.
Given: $\kappa = 1.5 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$ and $R = 150 \ \Omega$.
Substituting the values: $1.5 \times 10^{-4} = \frac{1}{150} \times G^*$.
Therefore,$G^* = 1.5 \times 10^{-4} \times 150 = 0.0225 \ cm^{-1}$.
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242
ChemistryEasyMCQMHT CET · 2022
$A$ conductivity cell has two electrodes $18 \ mm$ apart and having cross-sectional area $2.0 \ cm^2$. What is the value of the cell constant (in $cm^{-1}$)?
A
$0.9$
B
$0.18$
C
$3.6$
D
$0.2$
Solution
(A) The cell constant $(G^*)$ is defined as the ratio of the distance between electrodes $(L)$ to the cross-sectional area $(A)$.
Given: $L = 18 \ mm = 1.8 \ cm$ and $A = 2.0 \ cm^2$.
Cell constant $= \frac{L}{A} = \frac{1.8 \ cm}{2.0 \ cm^2} = 0.9 \ cm^{-1}$.
Given: $L = 18 \ mm = 1.8 \ cm$ and $A = 2.0 \ cm^2$.
Cell constant $= \frac{L}{A} = \frac{1.8 \ cm}{2.0 \ cm^2} = 0.9 \ cm^{-1}$.
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243
ChemistryMediumMCQMHT CET · 2022
What is the value of conductivity of $0.01 \ M$ solution of an electrolyte having molar conductivity $141 \ \Omega^{-1} \ cm^2 \ mol^{-1}$?
A
$5.64 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
B
$1.41 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
C
$4.23 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
D
$7.09 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
Solution
(B) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(K)$ is given by the formula:
$\Lambda_m = \frac{K \times 1000}{M}$
Rearranging the formula to solve for conductivity $(K)$:
$K = \frac{\Lambda_m \times M}{1000}$
Given:
$\Lambda_m = 141 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$M = 0.01 \ M$
Substituting the values:
$K = \frac{141 \times 0.01}{1000} = \frac{1.41}{1000} = 1.41 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
$\Lambda_m = \frac{K \times 1000}{M}$
Rearranging the formula to solve for conductivity $(K)$:
$K = \frac{\Lambda_m \times M}{1000}$
Given:
$\Lambda_m = 141 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$M = 0.01 \ M$
Substituting the values:
$K = \frac{141 \times 0.01}{1000} = \frac{1.41}{1000} = 1.41 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
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244
ChemistryMediumMCQMHT CET · 2022
What is the $SI$ unit of conductivity?
A
$S \ m^{-1}$
B
$S \ m^2 \ mol^{-1}$
C
$S \ cm^{-1}$
D
$S \ m \ mol^{-1}$
Solution
(A) Conductivity $(\kappa)$ is defined as the reciprocal of resistivity $(\rho)$.
Since the $SI$ unit of resistivity is $\Omega \ m$,the $SI$ unit of conductivity is $\Omega^{-1} \ m^{-1}$.
$\Omega^{-1}$ is also known as Siemens $(S)$.
Therefore,the $SI$ unit of conductivity is $S \ m^{-1}$.
Since the $SI$ unit of resistivity is $\Omega \ m$,the $SI$ unit of conductivity is $\Omega^{-1} \ m^{-1}$.
$\Omega^{-1}$ is also known as Siemens $(S)$.
Therefore,the $SI$ unit of conductivity is $S \ m^{-1}$.
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245
ChemistryMediumMCQMHT CET · 2022
Molar conductivity of $0.01 \ M \ CH_3COOH$ is $19.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. Calculate its degree of dissociation if molar conductivity at zero concentration is $390 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
A
$0.05$
B
$0.2$
C
$0.08$
D
$0.6$
Solution
(A) The degree of dissociation $(\alpha)$ is given by the ratio of molar conductivity at a given concentration $(\Lambda_m)$ to the molar conductivity at infinite dilution $(\Lambda_m^{\circ})$.
$\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$
Given,$\Lambda_m = 19.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $\Lambda_m^{\circ} = 390 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$\alpha = \frac{19.5}{390} = 0.05$.
$\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$
Given,$\Lambda_m = 19.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $\Lambda_m^{\circ} = 390 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$\alpha = \frac{19.5}{390} = 0.05$.
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246
ChemistryEasyMCQMHT CET · 2022
Calculate the molar conductivity of $CH_2ClCOOH$ at zero concentration if molar conductivities of $HCl$,$KCl$,and $CH_2ClCOOK$ at zero concentration are $4.2$,$1.4$,and $1.1 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.
A
$3.9 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
B
$4.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
C
$6.6 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
D
$1.7 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Solution
(A) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution (zero concentration) can be expressed as:
$\Lambda^{\circ}_{CH_2ClCOOH} = \Lambda^{\circ}_{CH_2ClCOOK} + \Lambda^{\circ}_{HCl} - \Lambda^{\circ}_{KCl}$
Substituting the given values:
$\Lambda^{\circ}_{CH_2ClCOOH} = 1.1 + 4.2 - 1.4$
$\Lambda^{\circ}_{CH_2ClCOOH} = 3.9 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\Lambda^{\circ}_{CH_2ClCOOH} = \Lambda^{\circ}_{CH_2ClCOOK} + \Lambda^{\circ}_{HCl} - \Lambda^{\circ}_{KCl}$
Substituting the given values:
$\Lambda^{\circ}_{CH_2ClCOOH} = 1.1 + 4.2 - 1.4$
$\Lambda^{\circ}_{CH_2ClCOOH} = 3.9 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
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247
ChemistryEasyMCQMHT CET · 2022
Calculate $\wedge_0$ of $CH_2ClCOOH$ if $\wedge_0$ for $HCl, KCl$ and $CH_2ClCOOK$ are $4.2, 1.5$ and $1.1 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.
A
$1.9 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
B
$4.2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
C
$2.7 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
D
$3.8 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Solution
(D) According to Kohlrausch's law of independent migration of ions:
$\wedge_0(CH_2ClCOOH) = \wedge_0(CH_2ClCOOK) + \wedge_0(HCl) - \wedge_0(KCl)$
Substituting the given values:
$\wedge_0(CH_2ClCOOH) = 1.1 + 4.2 - 1.5$
$\wedge_0(CH_2ClCOOH) = 3.8 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\wedge_0(CH_2ClCOOH) = \wedge_0(CH_2ClCOOK) + \wedge_0(HCl) - \wedge_0(KCl)$
Substituting the given values:
$\wedge_0(CH_2ClCOOH) = 1.1 + 4.2 - 1.5$
$\wedge_0(CH_2ClCOOH) = 3.8 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
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248
ChemistryEasyMCQMHT CET · 2022
Which among the following is an example of a $2$-dimensional nanostructure?
A
Microcapsules
B
Nanowires
C
Thin films
D
Nanorings
Solution
(C) Nanomaterials are classified based on their dimensions.
$0$-dimensional nanostructures have all three dimensions in the nanoscale (e.g.,nanoparticles).
$1$-dimensional nanostructures have two dimensions in the nanoscale (e.g.,nanowires,nanotubes).
$2$-dimensional nanostructures have one dimension in the nanoscale (e.g.,thin films,coatings).
Therefore,thin films are an example of a $2$-dimensional nanostructure.
$0$-dimensional nanostructures have all three dimensions in the nanoscale (e.g.,nanoparticles).
$1$-dimensional nanostructures have two dimensions in the nanoscale (e.g.,nanowires,nanotubes).
$2$-dimensional nanostructures have one dimension in the nanoscale (e.g.,thin films,coatings).
Therefore,thin films are an example of a $2$-dimensional nanostructure.
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249
ChemistryEasyMCQMHT CET · 2022
Which among the following compounds is a vinylic halide?
A
$1-$chloroethene
B
$2-$Bromo$-2-$methylpropane
C
$2-$chloropropane
D
Bromoethane
Solution
(A) vinylic halide is a compound in which the halogen atom is directly attached to a carbon atom that is part of a carbon-carbon double bond $(C=C-X)$.
In $1-$chloroethene $(CH_2=CHCl)$,the chlorine atom is attached to a carbon atom involved in a double bond,which satisfies the definition of a vinylic halide.
Therefore,the correct option is $A$.
In $1-$chloroethene $(CH_2=CHCl)$,the chlorine atom is attached to a carbon atom involved in a double bond,which satisfies the definition of a vinylic halide.
Therefore,the correct option is $A$.
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250
ChemistryMediumMCQMHT CET · 2022
Identify the simple (symmetrical) ether from the following.
A
Methoxyethane
B
Anisole
C
$1$-propoxybenzene
D
Methoxymethane
Solution
(D) simple or symmetrical ether is one in which both alkyl or aryl groups attached to the oxygen atom are the same.
In $CH_3-O-CH_3$ (Methoxymethane),both groups attached to the oxygen atom are methyl groups $(-CH_3)$.
Therefore,it is a simple ether.
In $CH_3-O-CH_3$ (Methoxymethane),both groups attached to the oxygen atom are methyl groups $(-CH_3)$.
Therefore,it is a simple ether.
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