Ncert Solutions Chemistry Class 12th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, ii)

Conductance is the reciprocal of resistance and conductivity is the reciprocal of resistivity. It can be written as follows

k = $\frac{1}{\mathbf{?}}$

R= $\mathbf{?}\mathbf{}\frac{1}{\mathbf{A}}$

$\mathbf{?}$ = $\frac{\mathbf{R}\mathbf{A}}{\mathbf{l}}$

k  = $\frac{1}{\frac{\mathbf{R}\mathbf{A}}{\mathbf{l}}}$

k = $\frac{1}{\mathbf{R}}\frac{\mathbf{l}}{\mathbf{A}}$

k = $\frac{1}{\mathbf{R}}$ G*

k = $\frac{{\mathbf{G}}^{\mathbf{*}}}{\mathbf{R}}$

G* is cell constant.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, ii)

In the electrolysis process of CuSO₄ the following reactions occur at the half cell which is as follows:

At the cathode, the reaction goes this way,

Cu²⁺+2e⁻→Cu (s)

At the anode, the reaction goes this way,

Cu (s)→Cu²⁺+2e−

Here copper will be deposited at the cathode and copper will dissolve at the anode.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Correct options: A and B

Arrhenius equation k = Ae – $\frac{{\mathbf{E}}_{\mathbf{a}}}{\mathbf{R}\mathbf{T}}$

When the temperature rises, $\frac{{\mathbf{E}}_{\mathbf{a}}}{\mathbf{R}\mathbf{T}}$ value falls and – $\frac{{\mathbf{E}}_{\mathbf{a}}}{\mathbf{R}\mathbf{T}}$

As a result, e- $\frac{{\mathbf{E}}_{\mathbf{a}}}{\mathbf{R}\mathbf{T}}$  increases, and it is certain that k and  T increases as well. As a result, Option A is proven.

Catalyst increases the rate of reaction by lowering the activation energy. Thus option B is also correct. 

This is a Multiple Choice Questions as classified in NCERT Exemplar

Correct options: A and D

According to the graph, the area under the cure must not vary as the temperature rises.T2 > T1

The energy level of T2 is higher than T1.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, iii)

Electrolysis of copper sulfate solution is as follows

CuSO₄ ? Cu²⁺ + SO₄²⁻

H₂O? H⁺ + OH⁻

At the cathode, the reaction goes this way,

Cu²⁺ + 2e− → Cu; E? _Cell = 0.34V

H₂O⁻ → $\frac{1}{2}$ H₂ E? _Cell = 0.00V

At the anode, the reaction goes this way,

2SO₄^2- + 2e−→S₂O₈²⁻ E? _Cell =1.96V

2H₂O→O₂ + 4H⁺ + 4e⁻ E? _Cell =1.23V

The reaction will lower the value of E? _Cell is preferred at anode so the second reaction is feasible.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, iii)

 Kohlrausch law of states that limiting molar conductivity of any salt species is equal to the sum of the limiting molar conductivity of cations and anions of the electrolyte

This is a Multiple Choice Questions as classified in NCERT Exemplar

Correct options: A and C

The Maxwell Boltzmann distribution, often known as the Gibbs distribution, is a measure that expresses the likelihood of a system being in each state as a function of the energy of that state.

Look at the graph.

T₂ > T₁

The fraction of molecules falls as the temperature rises because the area under the curve shrinks.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, ii)

Conductivity is due to the movement of ions in the solution. The conductivity of ions depends on the following factors:

(i) nature of electrolyte added

(ii) size of ion produced

(iii) concentration of electrolyte

(iv) nature of the solvent

(v) temperature

Distance between electrodes does not affect the conductivity of an electrolytic solution.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (ii, iii)

At equilibrium, ?G $°$  = -2.303 RT ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

-nFE $°$  = -2.303 RT ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

E $°$ = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{\mathrm{n}\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

For Daniel cell, n=2

E $°$ = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

At equilibrium, E $°$ = 1.1V

1.1V = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$

(i) Since ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$  this option is incorrect

(ii) As derived 1.1V = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

(iii) As derived _{${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$} so this option is correct

(iv) As ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$  _,so this option is incorrect

This is a Multiple Choice Questions as classified in NCERT Exemplar

Correct options: A and D

Activation Energy is the amount of energy released when reactant molecules collide and form an activated complex. When energy is released, the complex decomposes into a product.