ChemistryQ201–213 of 236 questions
Page 5 of 5 · English
201
ChemistryEasyMCQTS EAMCET · 2001
What is the molecular formula of borazole?
A
$B_2H_6$
B
$B_6N_6H_6$
C
$B_3N_3H_6$
D
$B_3N_3H_3$
Solution
(C) Borazole,also known as inorganic benzene,has the chemical formula $B_3N_3H_6$.
It is isoelectronic with benzene $(C_6H_6)$ and possesses a cyclic structure consisting of alternating boron and nitrogen atoms.
It is isoelectronic with benzene $(C_6H_6)$ and possesses a cyclic structure consisting of alternating boron and nitrogen atoms.
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202
ChemistryMediumMCQTS EAMCET · 2001
The catenation tendency of $C, Si$ and $Ge$ is in the order $Ge < Si < C$. The bond energies (in $kJ \ mol^{-1}$) of $C-C, Si-Si$ and $Ge-Ge$ bonds,respectively,are:
A
$167, 180, 348$
B
$180, 167, 348$
C
$348, 167, 180$
D
$348, 180, 167$
Solution
(D) The catenation tendency is directly proportional to the bond dissociation energy of the element-element bond.
As the atomic size increases down the group,the bond length increases and the bond dissociation energy decreases.
The bond energies for $C-C$,$Si-Si$,and $Ge-Ge$ are approximately $348 \ kJ \ mol^{-1}$,$180 \ kJ \ mol^{-1}$,and $167 \ kJ \ mol^{-1}$ respectively.
Therefore,the order of bond energies is $C-C > Si-Si > Ge-Ge$.
As the atomic size increases down the group,the bond length increases and the bond dissociation energy decreases.
The bond energies for $C-C$,$Si-Si$,and $Ge-Ge$ are approximately $348 \ kJ \ mol^{-1}$,$180 \ kJ \ mol^{-1}$,and $167 \ kJ \ mol^{-1}$ respectively.
Therefore,the order of bond energies is $C-C > Si-Si > Ge-Ge$.
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203
ChemistryMediumMCQTS EAMCET · 2001
Which one of the following is used in the preparation of styrene?
A
$CH_3CHO$
B
$P_2O_5$
C
$CH_4$
D
$C_6H_6$
Solution
(D) Styrene $(C_6H_5CH=CH_2)$ is prepared from benzene $(C_6H_6)$ in two steps:
$1$. Ethylbenzene is prepared by the alkylation of benzene with ethene in the presence of an $AlCl_3$ catalyst: $C_6H_6 + CH_2=CH_2 \xrightarrow{AlCl_3} C_6H_5CH_2CH_3$.
$2$. Ethylbenzene is then dehydrogenated at high temperature $(700^{\circ}C)$ using a catalyst (typically iron oxide or $Al_2O_3$) to produce styrene: $C_6H_5CH_2CH_3 \xrightarrow{700^{\circ}C, \text{catalyst}} C_6H_5CH=CH_2 + H_2$.
$1$. Ethylbenzene is prepared by the alkylation of benzene with ethene in the presence of an $AlCl_3$ catalyst: $C_6H_6 + CH_2=CH_2 \xrightarrow{AlCl_3} C_6H_5CH_2CH_3$.
$2$. Ethylbenzene is then dehydrogenated at high temperature $(700^{\circ}C)$ using a catalyst (typically iron oxide or $Al_2O_3$) to produce styrene: $C_6H_5CH_2CH_3 \xrightarrow{700^{\circ}C, \text{catalyst}} C_6H_5CH=CH_2 + H_2$.
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204
ChemistryMediumMCQTS EAMCET · 2001
In organic reactions,sodium in liquid ammonia is used as:
A
reducing agent
B
hydrating agent
C
oxidising agent
D
precipitating agent
Solution
(A) Sodium in liquid ammonia $(Na/NH_3(l))$ is a well-known reagent used for the Birch reduction.
In this reaction,the sodium metal dissolves in liquid ammonia to release solvated electrons,which act as a powerful $reducing \text{ } agent$.
It is specifically used to reduce alkynes to trans-alkenes and aromatic rings to $1,4-$cyclohexadienes.
In this reaction,the sodium metal dissolves in liquid ammonia to release solvated electrons,which act as a powerful $reducing \text{ } agent$.
It is specifically used to reduce alkynes to trans-alkenes and aromatic rings to $1,4-$cyclohexadienes.
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205
ChemistryMediumMCQTS EAMCET · 2001
What are the oxidation numbers of nitrogen in $NH_4NO_3$?
A
$+3, -5$
B
$-3, +5$
C
$+3, -6$
D
$+2, +2$
Solution
(B) $NH_4NO_3$ consists of $NH_4^+$ and $NO_3^-$ ions.
For $NH_4^+$: Let the oxidation number of $N$ be $x$. $x + 4(+1) = +1$,so $x = -3$.
For $NO_3^-$: Let the oxidation number of $N$ be $x$. $x + 3(-2) = -1$,so $x - 6 = -1$,which gives $x = +5$.
Thus,the oxidation numbers of nitrogen are $-3$ and $+5$.
For $NH_4^+$: Let the oxidation number of $N$ be $x$. $x + 4(+1) = +1$,so $x = -3$.
For $NO_3^-$: Let the oxidation number of $N$ be $x$. $x + 3(-2) = -1$,so $x - 6 = -1$,which gives $x = +5$.
Thus,the oxidation numbers of nitrogen are $-3$ and $+5$.
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206
ChemistryDifficultMCQTS EAMCET · 2001
How many litres of oxygen (at $STP$) are required for complete combustion of $39 \ g$ of liquid benzene?
(Atomic weights: $C=12, O=16, H=1$)
(Atomic weights: $C=12, O=16, H=1$)
A
$84$
B
$22.4$
C
$42$
D
$11.2$
Solution
(A) The balanced chemical equation for the combustion of benzene $(C_6H_6)$ is:
$2C_6H_6(l) + 15O_2(g) \longrightarrow 12CO_2(g) + 6H_2O(l)$
Calculate the molar mass of benzene $(C_6H_6)$:
$M = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \ g/mol$
From the stoichiometry of the reaction,$2 \ mol$ of $C_6H_6$ require $15 \ mol$ of $O_2$.
Therefore,$78 \ g$ of $C_6H_6$ $(1 \ mol)$ requires $7.5 \ mol$ of $O_2$.
At $STP$,$1 \ mol$ of gas occupies $22.4 \ L$.
Volume of $O_2$ required for $78 \ g$ of $C_6H_6 = 7.5 \times 22.4 \ L = 168 \ L$.
For $39 \ g$ of $C_6H_6$ (which is $0.5 \ mol$):
Volume of $O_2 = \frac{168 \ L}{2} = 84 \ L$.
$2C_6H_6(l) + 15O_2(g) \longrightarrow 12CO_2(g) + 6H_2O(l)$
Calculate the molar mass of benzene $(C_6H_6)$:
$M = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \ g/mol$
From the stoichiometry of the reaction,$2 \ mol$ of $C_6H_6$ require $15 \ mol$ of $O_2$.
Therefore,$78 \ g$ of $C_6H_6$ $(1 \ mol)$ requires $7.5 \ mol$ of $O_2$.
At $STP$,$1 \ mol$ of gas occupies $22.4 \ L$.
Volume of $O_2$ required for $78 \ g$ of $C_6H_6 = 7.5 \times 22.4 \ L = 168 \ L$.
For $39 \ g$ of $C_6H_6$ (which is $0.5 \ mol$):
Volume of $O_2 = \frac{168 \ L}{2} = 84 \ L$.
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207
ChemistryDifficultMCQTS EAMCET · 2001
What is the volume (in litres) of oxygen at $STP$ required for complete combustion of $32 \,g$ of $CH_4$? (Molecular weight of $CH_4$ is $16$)
A
$44.8$
B
$89.6$
C
$22.4$
D
$179.2$
Solution
(B) The balanced chemical equation for the combustion of methane is:
$CH_4(g) + 2O_2(g) \longrightarrow CO_2(g) + 2H_2O(l)$
From the stoichiometry,$1 \,mol$ of $CH_4$ $(16 \,g)$ requires $2 \,mol$ of $O_2$.
At $STP$,$1 \,mol$ of gas occupies $22.4 \,L$.
Therefore,$16 \,g$ of $CH_4$ requires $2 \times 22.4 \,L = 44.8 \,L$ of $O_2$.
For $32 \,g$ of $CH_4$,the volume of $O_2$ required is:
$\frac{44.8 \,L}{16 \,g} \times 32 \,g = 89.6 \,L$.
$CH_4(g) + 2O_2(g) \longrightarrow CO_2(g) + 2H_2O(l)$
From the stoichiometry,$1 \,mol$ of $CH_4$ $(16 \,g)$ requires $2 \,mol$ of $O_2$.
At $STP$,$1 \,mol$ of gas occupies $22.4 \,L$.
Therefore,$16 \,g$ of $CH_4$ requires $2 \times 22.4 \,L = 44.8 \,L$ of $O_2$.
For $32 \,g$ of $CH_4$,the volume of $O_2$ required is:
$\frac{44.8 \,L}{16 \,g} \times 32 \,g = 89.6 \,L$.
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208
ChemistryDifficultMCQTS EAMCET · 2001
$250 \ mL$ of a sodium carbonate solution contains $2.65 \ g$ of $Na_2CO_3$. If $10 \ mL$ of this solution is diluted to $1 \ L$,what is the concentration of the resultant solution (in $M$)?
A
$0.1$
B
$0.01$
C
$0.001$
D
$0.02$
Solution
(C) Step $1$: Calculate the molarity of the initial $250 \ mL$ solution.
$M = \frac{\text{mass of solute}}{\text{molar mass}} \times \frac{1000}{V(mL)}$
$M = \frac{2.65 \ g}{106 \ g/mol} \times \frac{1000}{250 \ mL} = 0.1 \ M$
Step $2$: Use the dilution formula $M_1V_1 = M_2V_2$ to find the concentration of the diluted solution.
$M_1 = 0.1 \ M, V_1 = 10 \ mL$
$V_2 = 1 \ L = 1000 \ mL$
$0.1 \times 10 = M_2 \times 1000$
$M_2 = \frac{0.1 \times 10}{1000} = 0.001 \ M$
$M = \frac{\text{mass of solute}}{\text{molar mass}} \times \frac{1000}{V(mL)}$
$M = \frac{2.65 \ g}{106 \ g/mol} \times \frac{1000}{250 \ mL} = 0.1 \ M$
Step $2$: Use the dilution formula $M_1V_1 = M_2V_2$ to find the concentration of the diluted solution.
$M_1 = 0.1 \ M, V_1 = 10 \ mL$
$V_2 = 1 \ L = 1000 \ mL$
$0.1 \times 10 = M_2 \times 1000$
$M_2 = \frac{0.1 \times 10}{1000} = 0.001 \ M$
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209
ChemistryMediumMCQTS EAMCET · 2001
$n$ moles of an ideal gas at temperature $T$ (in kelvin) occupy $V \ L$ of volume,exerting a pressure of $P$ atmospheres. What is the concentration (in $mol / L$)?
A
$\frac{P}{RT}$
B
$\frac{PT}{R}$
C
$\frac{RT}{P}$
D
$\frac{R}{PT}$
Solution
(A) The ideal gas equation is given by $PV = nRT$.
Concentration is defined as the number of moles per unit volume,i.e.,$C = \frac{n}{V}$.
From the ideal gas equation,we can rearrange the terms to find $\frac{n}{V}$:
$\frac{n}{V} = \frac{P}{RT}$.
Therefore,the concentration is $\frac{P}{RT} \ mol/L$.
Concentration is defined as the number of moles per unit volume,i.e.,$C = \frac{n}{V}$.
From the ideal gas equation,we can rearrange the terms to find $\frac{n}{V}$:
$\frac{n}{V} = \frac{P}{RT}$.
Therefore,the concentration is $\frac{P}{RT} \ mol/L$.
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210
ChemistryDifficultMCQTS EAMCET · 2001
The r.m.s. velocity of $CO_2$ at a temperature $T$ (in kelvin) is $x \ cm \ s^{-1}$. At what temperature (in kelvin),the r.m.s. velocity of nitrous oxide would be $4 x \ cm \ s^{-1}$ (in $T$)? (Atomic weights of $C, N$ and $O$ are respectively $12, 14$ and $16$)
A
$16$
B
$2$
C
$4$
D
$32$
Solution
(A) The formula for r.m.s. velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
For $CO_2$: $M_{CO_2} = 12 + (2 \times 16) = 44 \ g \ mol^{-1}$. Given $v_{rms} = x$ at temperature $T$.
For $N_2O$: $M_{N_2O} = (2 \times 14) + 16 = 44 \ g \ mol^{-1}$. Let the temperature be $T'$. Given $v_{rms} = 4x$.
Since $v_{rms} \propto \sqrt{T/M}$ and $M_{CO_2} = M_{N_2O} = 44$,we have $\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.
Substituting the values: $\frac{x}{4x} = \sqrt{\frac{T}{T'}}$.
$\frac{1}{4} = \sqrt{\frac{T}{T'}}$.
Squaring both sides: $\frac{1}{16} = \frac{T}{T'}$.
Therefore,$T' = 16T$.
For $CO_2$: $M_{CO_2} = 12 + (2 \times 16) = 44 \ g \ mol^{-1}$. Given $v_{rms} = x$ at temperature $T$.
For $N_2O$: $M_{N_2O} = (2 \times 14) + 16 = 44 \ g \ mol^{-1}$. Let the temperature be $T'$. Given $v_{rms} = 4x$.
Since $v_{rms} \propto \sqrt{T/M}$ and $M_{CO_2} = M_{N_2O} = 44$,we have $\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.
Substituting the values: $\frac{x}{4x} = \sqrt{\frac{T}{T'}}$.
$\frac{1}{4} = \sqrt{\frac{T}{T'}}$.
Squaring both sides: $\frac{1}{16} = \frac{T}{T'}$.
Therefore,$T' = 16T$.
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211
ChemistryMediumMCQTS EAMCET · 2001
The energy of an electron present in Bohr's second orbit of hydrogen atom is
A
$-1312 \ J \ \text{atom}^{-1}$
B
$-328 \ kJ \ mol^{-1}$
C
$-328 \ J \ mol^{-1}$
D
$-164 \ kJ \ mol^{-1}$
Solution
(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = \frac{-1312}{n^2} \ kJ \ mol^{-1}$.
For the second orbit,$n = 2$.
Substituting the value of $n$ in the formula:
$E_2 = \frac{-1312}{2^2} \ kJ \ mol^{-1} = \frac{-1312}{4} \ kJ \ mol^{-1} = -328 \ kJ \ mol^{-1}$.
For the second orbit,$n = 2$.
Substituting the value of $n$ in the formula:
$E_2 = \frac{-1312}{2^2} \ kJ \ mol^{-1} = \frac{-1312}{4} \ kJ \ mol^{-1} = -328 \ kJ \ mol^{-1}$.
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212
ChemistryMediumMCQTS EAMCET · 2001
In the ground state,an element has $13$ electrons in the $M$ shell. The element is
A
copper
B
chromium
C
nickel
D
iron
Solution
(B) The $M$ shell corresponds to the $n=3$ principal energy level.
For an element to have $13$ electrons in the $M$ shell,the electronic configuration must include $3s^2, 3p^6, 3d^5$.
Considering the Aufbau principle and stability,the configuration is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$.
The total number of electrons is $2+2+6+2+6+5+1 = 24$.
The element with atomic number $24$ is chromium $(Cr)$.
For an element to have $13$ electrons in the $M$ shell,the electronic configuration must include $3s^2, 3p^6, 3d^5$.
Considering the Aufbau principle and stability,the configuration is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$.
The total number of electrons is $2+2+6+2+6+5+1 = 24$.
The element with atomic number $24$ is chromium $(Cr)$.
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213
ChemistryMCQTS EAMCET · 2001
$A$ steel meter scale is to be ruled so that millimeter intervals are accurate within about $5 \times 10^{-5} \ m$ at a certain temperature. The maximum temperature variation allowable during the ruling is (Coefficient of linear expansion of steel $= 10 \times 10^{-6} \ K^{-1}$) (in $^{\circ} C$)
A
$2$
B
$5$
C
$7$
D
$10$
Solution
(B) The formula for linear thermal expansion is given by $\Delta L = L \alpha \Delta T$,where $\Delta L$ is the change in length,$L$ is the original length,$\alpha$ is the coefficient of linear expansion,and $\Delta T$ is the change in temperature.
Given:
$\Delta L = 5 \times 10^{-5} \ m$
$L = 1 \ m$
$\alpha = 10 \times 10^{-6} \ K^{-1}$
Rearranging the formula to solve for $\Delta T$:
$\Delta T = \frac{\Delta L}{L \alpha}$
Substituting the values:
$\Delta T = \frac{5 \times 10^{-5}}{1 \times 10 \times 10^{-6}}$
$\Delta T = \frac{5 \times 10^{-5}}{10^{-5}}$
$\Delta T = 5^{\circ} C$
Therefore,the maximum temperature variation allowable is $5^{\circ} C$.
Given:
$\Delta L = 5 \times 10^{-5} \ m$
$L = 1 \ m$
$\alpha = 10 \times 10^{-6} \ K^{-1}$
Rearranging the formula to solve for $\Delta T$:
$\Delta T = \frac{\Delta L}{L \alpha}$
Substituting the values:
$\Delta T = \frac{5 \times 10^{-5}}{1 \times 10 \times 10^{-6}}$
$\Delta T = \frac{5 \times 10^{-5}}{10^{-5}}$
$\Delta T = 5^{\circ} C$
Therefore,the maximum temperature variation allowable is $5^{\circ} C$.
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