QBManager

ELECTROCHEMISTRY

276136 Unit of equivalent conductance is :

1

{ohm}^{- 1} {cm}^{2} (g - eq)^{- 1}

2

ohm cm (g - eq)

3

ohm cm^{2} (g - eq)^{- 1}

4

{ohm}^{- 1} cm (g - eq)^{- 1}

SRM JEE 2014

Explanation:

The equivalent conductance is calculated by multiplying the specific conductivity and the volume of the solution having one gram equivalent of the electrolyte.
$\begin{aligned} Equivalent conductivity = κ \times \frac{1000}{Normality} \\ = \frac{{ohm}^{- 1} cm}{g_{equivalent {cm}^{- 1}}} \\ {= {ohm}^{- 1} {cm}^{2} (g equivalent)}^{- 1} \\ = {ohm}^{- 1} {cm}^{2} (g - eq)^{- 1} \end{aligned}$

2006

ELECTROCHEMISTRY

276138 Equivalent conductance's of ${Ba}^{2 +}$ and ${Cl}^{-}$ ions are 127 and $76 {ohm}^{- 1} {cm}^{- 1} {eq}^{- 1}$ respectively. Equivalent conductance of ${BaCl}_{2}$ at infinite dilution is

1 139.5

2 101.5

3 203

4 279

Explanation:

Chemical equation-
${BaCl}_{2} \to {Ba}^{2 +} + 2 {Cl}^{-}$
From the Kohlrausch law -*
$\begin{aligned} λ_{\infty} = \frac{1}{n} λ_{+}^{\infty} + \frac{1}{n} λ_{-}^{\infty} \\ λ_{\infty} ({BaCl}_{2}) = \frac{1}{2} \times λ_{{Ba}^{2 +}}^{\infty} + \frac{1}{1} \times λ_{{Cl}^{-}}^{\infty} \\ = \frac{127}{2} + 76 \\ = 63.5 + 76 \\ = 139.5 {ohm}^{- 1} {cm}^{- 1} {eq}^{- 1} \end{aligned}$

ELECTROCHEMISTRY

276133 Resistance of a conductivity cell filled with 0.1 mol $L^{- 1} NaCl$ is $100 Ohm$ . If the resistance of the same cell when filled with $0.02 mol L^{- 1} NaCl$ solution is $258 Ohm$ . The conductivity of $0.02 mol L^{- 1} NaCl$ solution is (Conductivity of $0.1 L^{- 1} NaCl$ is $1.29 {Sm}^{- 1}$ )

1

1.0 S m^{- 1}

2

0.2 S m^{- 1}

3

2.0 S m^{- 1}

4

0.5 S m^{- 1}

Explanation:

For $0.1 mol L^{- 1} NaCl$ solution,
Conductivity $(K) = 1.29 S m^{- 1}$ , Resistance $(R) = 100 Ω$
$∵$ Cell constant $=$ Conductivity $\times$ Resistance
$= 1.29 \times 100 = 129$
For, $0.02 mol L^{- 1} NaCl$ solution
Resistance $= 258 Ω$
$\begin{aligned} Conductivity (K) & = \frac{Cell constant}{Resistance} \\ = \frac{129}{258} \\ = 0.5 S m^{- 1} \end{aligned}$

AP-EAMCET-05.07.2022

ELECTROCHEMISTRY

276134 At $25^{\circ} C$ , the molar conductances at infinite dilution for the strong electrolytes $NaOH, NaCl$ and ${BaCl}_{2}$ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times$ $10^{- 4} {Sm}^{2} {mol}^{- 1}$ respectively, $λ_{m}^{\circ} Ba (OH)_{2}$ in ${Sm}^{2}$ mol $^{- 1}$ is

1

52.4 \times 10^{- 4}

2

524 \times 10^{- 4}

3

402 \times 10^{- 4}

4

262 \times 10^{- 4}

Explanation:

$\begin{aligned} Given that, \\ {BaCl}_{2} = 280 \times 10^{- 4} \\ NaOH = 248 \times 10^{- 4} \\ NaCl = 126 \times 10^{- 4} \\ {BaCl}_{2} + 2 NaOH ⟶ Ba (OH)_{2} + 2 NaCl \\ λ_{mBa (OH)_{2}}^{\infty} = λ_{mBaCl}^{2} + 2 λ_{m NaOH}^{\infty} - 2 λ_{m NaCl}^{\infty} \\ = 280 \times 10^{- 4} + 2 \times 248 \times 10^{- 4} - 2 \times 126 \times 10^{- 4} \\ = (280 + 496 - 252) \times 10^{- 4} \\ = 524 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \end{aligned}$

ELECTROCHEMISTRY

276135 The molar conductance of $NaCl, HCl$ and ${CH}_{3} COONa$ at infinite dilution are 126.45, 426.16 and $91.0 S {cm}^{2} {mol}^{- 1}$ respectively. The molar conductance of ${CH}_{3} COOH$ at infinte dilution is Choose the right option for your answer.

1

540.48 S {cm}^{2} {mol}^{- 1}

2

201.28 {Scm}^{2} {mol}^{- 1}

3

390.71 S {cm}^{2} {mol}^{- 1}

4

698.28 {Scm}^{2} {mol}^{- 1}

Explanation:

Given that,
$\begin{aligned} Λ_{m}^{\circ} NaCl = 126.5 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} HCl = 426.16 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} {CH}_{3} COONa = 91.0 S {cm}^{2} {mol}^{- 1} \end{aligned}$
$Λ_{m}^{\circ} {CH}_{3} COOH = Λ_{m}^{\circ} {CH}_{3} COONa + Λ_{m}^{\circ} HCl - Λ_{m}^{\circ} NaCl$
$\begin{aligned} = 91.0 + 426.16 - 126.5 \\ = 390.71 S {cm}^{2} {mol}^{- 1} \end{aligned}$

MHT CET-03.05.2019

ELECTROCHEMISTRY

276136 Unit of equivalent conductance is :

1

{ohm}^{- 1} {cm}^{2} (g - eq)^{- 1}

2

ohm cm (g - eq)

3

ohm cm^{2} (g - eq)^{- 1}

4

{ohm}^{- 1} cm (g - eq)^{- 1}

SRM JEE 2014

Explanation:

The equivalent conductance is calculated by multiplying the specific conductivity and the volume of the solution having one gram equivalent of the electrolyte.
$\begin{aligned} Equivalent conductivity = κ \times \frac{1000}{Normality} \\ = \frac{{ohm}^{- 1} cm}{g_{equivalent {cm}^{- 1}}} \\ {= {ohm}^{- 1} {cm}^{2} (g equivalent)}^{- 1} \\ = {ohm}^{- 1} {cm}^{2} (g - eq)^{- 1} \end{aligned}$

2006

ELECTROCHEMISTRY

276138 Equivalent conductance's of ${Ba}^{2 +}$ and ${Cl}^{-}$ ions are 127 and $76 {ohm}^{- 1} {cm}^{- 1} {eq}^{- 1}$ respectively. Equivalent conductance of ${BaCl}_{2}$ at infinite dilution is

1 139.5

2 101.5

3 203

4 279

Explanation:

Chemical equation-
${BaCl}_{2} \to {Ba}^{2 +} + 2 {Cl}^{-}$
From the Kohlrausch law -*
$\begin{aligned} λ_{\infty} = \frac{1}{n} λ_{+}^{\infty} + \frac{1}{n} λ_{-}^{\infty} \\ λ_{\infty} ({BaCl}_{2}) = \frac{1}{2} \times λ_{{Ba}^{2 +}}^{\infty} + \frac{1}{1} \times λ_{{Cl}^{-}}^{\infty} \\ = \frac{127}{2} + 76 \\ = 63.5 + 76 \\ = 139.5 {ohm}^{- 1} {cm}^{- 1} {eq}^{- 1} \end{aligned}$

ELECTROCHEMISTRY

276133 Resistance of a conductivity cell filled with 0.1 mol $L^{- 1} NaCl$ is $100 Ohm$ . If the resistance of the same cell when filled with $0.02 mol L^{- 1} NaCl$ solution is $258 Ohm$ . The conductivity of $0.02 mol L^{- 1} NaCl$ solution is (Conductivity of $0.1 L^{- 1} NaCl$ is $1.29 {Sm}^{- 1}$ )

1

1.0 S m^{- 1}

2

0.2 S m^{- 1}

3

2.0 S m^{- 1}

4

0.5 S m^{- 1}

Explanation:

For $0.1 mol L^{- 1} NaCl$ solution,
Conductivity $(K) = 1.29 S m^{- 1}$ , Resistance $(R) = 100 Ω$
$∵$ Cell constant $=$ Conductivity $\times$ Resistance
$= 1.29 \times 100 = 129$
For, $0.02 mol L^{- 1} NaCl$ solution
Resistance $= 258 Ω$
$\begin{aligned} Conductivity (K) & = \frac{Cell constant}{Resistance} \\ = \frac{129}{258} \\ = 0.5 S m^{- 1} \end{aligned}$

AP-EAMCET-05.07.2022

ELECTROCHEMISTRY

276134 At $25^{\circ} C$ , the molar conductances at infinite dilution for the strong electrolytes $NaOH, NaCl$ and ${BaCl}_{2}$ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times$ $10^{- 4} {Sm}^{2} {mol}^{- 1}$ respectively, $λ_{m}^{\circ} Ba (OH)_{2}$ in ${Sm}^{2}$ mol $^{- 1}$ is

1

52.4 \times 10^{- 4}

2

524 \times 10^{- 4}

3

402 \times 10^{- 4}

4

262 \times 10^{- 4}

Explanation:

$\begin{aligned} Given that, \\ {BaCl}_{2} = 280 \times 10^{- 4} \\ NaOH = 248 \times 10^{- 4} \\ NaCl = 126 \times 10^{- 4} \\ {BaCl}_{2} + 2 NaOH ⟶ Ba (OH)_{2} + 2 NaCl \\ λ_{mBa (OH)_{2}}^{\infty} = λ_{mBaCl}^{2} + 2 λ_{m NaOH}^{\infty} - 2 λ_{m NaCl}^{\infty} \\ = 280 \times 10^{- 4} + 2 \times 248 \times 10^{- 4} - 2 \times 126 \times 10^{- 4} \\ = (280 + 496 - 252) \times 10^{- 4} \\ = 524 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \end{aligned}$

ELECTROCHEMISTRY

276135 The molar conductance of $NaCl, HCl$ and ${CH}_{3} COONa$ at infinite dilution are 126.45, 426.16 and $91.0 S {cm}^{2} {mol}^{- 1}$ respectively. The molar conductance of ${CH}_{3} COOH$ at infinte dilution is Choose the right option for your answer.

1

540.48 S {cm}^{2} {mol}^{- 1}

2

201.28 {Scm}^{2} {mol}^{- 1}

3

390.71 S {cm}^{2} {mol}^{- 1}

4

698.28 {Scm}^{2} {mol}^{- 1}

Explanation:

Given that,
$\begin{aligned} Λ_{m}^{\circ} NaCl = 126.5 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} HCl = 426.16 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} {CH}_{3} COONa = 91.0 S {cm}^{2} {mol}^{- 1} \end{aligned}$
$Λ_{m}^{\circ} {CH}_{3} COOH = Λ_{m}^{\circ} {CH}_{3} COONa + Λ_{m}^{\circ} HCl - Λ_{m}^{\circ} NaCl$
$\begin{aligned} = 91.0 + 426.16 - 126.5 \\ = 390.71 S {cm}^{2} {mol}^{- 1} \end{aligned}$

MHT CET-03.05.2019

ELECTROCHEMISTRY

276136 Unit of equivalent conductance is :

1

{ohm}^{- 1} {cm}^{2} (g - eq)^{- 1}

2

ohm cm (g - eq)

3

ohm cm^{2} (g - eq)^{- 1}

4

{ohm}^{- 1} cm (g - eq)^{- 1}

SRM JEE 2014

Explanation:

The equivalent conductance is calculated by multiplying the specific conductivity and the volume of the solution having one gram equivalent of the electrolyte.
$\begin{aligned} Equivalent conductivity = κ \times \frac{1000}{Normality} \\ = \frac{{ohm}^{- 1} cm}{g_{equivalent {cm}^{- 1}}} \\ {= {ohm}^{- 1} {cm}^{2} (g equivalent)}^{- 1} \\ = {ohm}^{- 1} {cm}^{2} (g - eq)^{- 1} \end{aligned}$

2006

ELECTROCHEMISTRY

276138 Equivalent conductance's of ${Ba}^{2 +}$ and ${Cl}^{-}$ ions are 127 and $76 {ohm}^{- 1} {cm}^{- 1} {eq}^{- 1}$ respectively. Equivalent conductance of ${BaCl}_{2}$ at infinite dilution is

1 139.5

2 101.5

3 203

4 279

Explanation:

Chemical equation-
${BaCl}_{2} \to {Ba}^{2 +} + 2 {Cl}^{-}$
From the Kohlrausch law -*
$\begin{aligned} λ_{\infty} = \frac{1}{n} λ_{+}^{\infty} + \frac{1}{n} λ_{-}^{\infty} \\ λ_{\infty} ({BaCl}_{2}) = \frac{1}{2} \times λ_{{Ba}^{2 +}}^{\infty} + \frac{1}{1} \times λ_{{Cl}^{-}}^{\infty} \\ = \frac{127}{2} + 76 \\ = 63.5 + 76 \\ = 139.5 {ohm}^{- 1} {cm}^{- 1} {eq}^{- 1} \end{aligned}$

ELECTROCHEMISTRY

276133 Resistance of a conductivity cell filled with 0.1 mol $L^{- 1} NaCl$ is $100 Ohm$ . If the resistance of the same cell when filled with $0.02 mol L^{- 1} NaCl$ solution is $258 Ohm$ . The conductivity of $0.02 mol L^{- 1} NaCl$ solution is (Conductivity of $0.1 L^{- 1} NaCl$ is $1.29 {Sm}^{- 1}$ )

1

1.0 S m^{- 1}

2

0.2 S m^{- 1}

3

2.0 S m^{- 1}

4

0.5 S m^{- 1}

Explanation:

For $0.1 mol L^{- 1} NaCl$ solution,
Conductivity $(K) = 1.29 S m^{- 1}$ , Resistance $(R) = 100 Ω$
$∵$ Cell constant $=$ Conductivity $\times$ Resistance
$= 1.29 \times 100 = 129$
For, $0.02 mol L^{- 1} NaCl$ solution
Resistance $= 258 Ω$
$\begin{aligned} Conductivity (K) & = \frac{Cell constant}{Resistance} \\ = \frac{129}{258} \\ = 0.5 S m^{- 1} \end{aligned}$

AP-EAMCET-05.07.2022

ELECTROCHEMISTRY

276134 At $25^{\circ} C$ , the molar conductances at infinite dilution for the strong electrolytes $NaOH, NaCl$ and ${BaCl}_{2}$ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times$ $10^{- 4} {Sm}^{2} {mol}^{- 1}$ respectively, $λ_{m}^{\circ} Ba (OH)_{2}$ in ${Sm}^{2}$ mol $^{- 1}$ is

1

52.4 \times 10^{- 4}

2

524 \times 10^{- 4}

3

402 \times 10^{- 4}

4

262 \times 10^{- 4}

Explanation:

$\begin{aligned} Given that, \\ {BaCl}_{2} = 280 \times 10^{- 4} \\ NaOH = 248 \times 10^{- 4} \\ NaCl = 126 \times 10^{- 4} \\ {BaCl}_{2} + 2 NaOH ⟶ Ba (OH)_{2} + 2 NaCl \\ λ_{mBa (OH)_{2}}^{\infty} = λ_{mBaCl}^{2} + 2 λ_{m NaOH}^{\infty} - 2 λ_{m NaCl}^{\infty} \\ = 280 \times 10^{- 4} + 2 \times 248 \times 10^{- 4} - 2 \times 126 \times 10^{- 4} \\ = (280 + 496 - 252) \times 10^{- 4} \\ = 524 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \end{aligned}$

ELECTROCHEMISTRY

276135 The molar conductance of $NaCl, HCl$ and ${CH}_{3} COONa$ at infinite dilution are 126.45, 426.16 and $91.0 S {cm}^{2} {mol}^{- 1}$ respectively. The molar conductance of ${CH}_{3} COOH$ at infinte dilution is Choose the right option for your answer.

1

540.48 S {cm}^{2} {mol}^{- 1}

2

201.28 {Scm}^{2} {mol}^{- 1}

3

390.71 S {cm}^{2} {mol}^{- 1}

4

698.28 {Scm}^{2} {mol}^{- 1}

Explanation:

Given that,
$\begin{aligned} Λ_{m}^{\circ} NaCl = 126.5 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} HCl = 426.16 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} {CH}_{3} COONa = 91.0 S {cm}^{2} {mol}^{- 1} \end{aligned}$
$Λ_{m}^{\circ} {CH}_{3} COOH = Λ_{m}^{\circ} {CH}_{3} COONa + Λ_{m}^{\circ} HCl - Λ_{m}^{\circ} NaCl$
$\begin{aligned} = 91.0 + 426.16 - 126.5 \\ = 390.71 S {cm}^{2} {mol}^{- 1} \end{aligned}$

MHT CET-03.05.2019

ELECTROCHEMISTRY

276136 Unit of equivalent conductance is :

1

{ohm}^{- 1} {cm}^{2} (g - eq)^{- 1}

2

ohm cm (g - eq)

3

ohm cm^{2} (g - eq)^{- 1}

4

{ohm}^{- 1} cm (g - eq)^{- 1}

SRM JEE 2014

Explanation:

The equivalent conductance is calculated by multiplying the specific conductivity and the volume of the solution having one gram equivalent of the electrolyte.
$\begin{aligned} Equivalent conductivity = κ \times \frac{1000}{Normality} \\ = \frac{{ohm}^{- 1} cm}{g_{equivalent {cm}^{- 1}}} \\ {= {ohm}^{- 1} {cm}^{2} (g equivalent)}^{- 1} \\ = {ohm}^{- 1} {cm}^{2} (g - eq)^{- 1} \end{aligned}$

2006

ELECTROCHEMISTRY

276138 Equivalent conductance's of ${Ba}^{2 +}$ and ${Cl}^{-}$ ions are 127 and $76 {ohm}^{- 1} {cm}^{- 1} {eq}^{- 1}$ respectively. Equivalent conductance of ${BaCl}_{2}$ at infinite dilution is

1 139.5

2 101.5

3 203

4 279

Explanation:

Chemical equation-
${BaCl}_{2} \to {Ba}^{2 +} + 2 {Cl}^{-}$
From the Kohlrausch law -*
$\begin{aligned} λ_{\infty} = \frac{1}{n} λ_{+}^{\infty} + \frac{1}{n} λ_{-}^{\infty} \\ λ_{\infty} ({BaCl}_{2}) = \frac{1}{2} \times λ_{{Ba}^{2 +}}^{\infty} + \frac{1}{1} \times λ_{{Cl}^{-}}^{\infty} \\ = \frac{127}{2} + 76 \\ = 63.5 + 76 \\ = 139.5 {ohm}^{- 1} {cm}^{- 1} {eq}^{- 1} \end{aligned}$

ELECTROCHEMISTRY

276133 Resistance of a conductivity cell filled with 0.1 mol $L^{- 1} NaCl$ is $100 Ohm$ . If the resistance of the same cell when filled with $0.02 mol L^{- 1} NaCl$ solution is $258 Ohm$ . The conductivity of $0.02 mol L^{- 1} NaCl$ solution is (Conductivity of $0.1 L^{- 1} NaCl$ is $1.29 {Sm}^{- 1}$ )

1

1.0 S m^{- 1}

2

0.2 S m^{- 1}

3

2.0 S m^{- 1}

4

0.5 S m^{- 1}

Explanation:

For $0.1 mol L^{- 1} NaCl$ solution,
Conductivity $(K) = 1.29 S m^{- 1}$ , Resistance $(R) = 100 Ω$
$∵$ Cell constant $=$ Conductivity $\times$ Resistance
$= 1.29 \times 100 = 129$
For, $0.02 mol L^{- 1} NaCl$ solution
Resistance $= 258 Ω$
$\begin{aligned} Conductivity (K) & = \frac{Cell constant}{Resistance} \\ = \frac{129}{258} \\ = 0.5 S m^{- 1} \end{aligned}$

AP-EAMCET-05.07.2022

ELECTROCHEMISTRY

276134 At $25^{\circ} C$ , the molar conductances at infinite dilution for the strong electrolytes $NaOH, NaCl$ and ${BaCl}_{2}$ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times$ $10^{- 4} {Sm}^{2} {mol}^{- 1}$ respectively, $λ_{m}^{\circ} Ba (OH)_{2}$ in ${Sm}^{2}$ mol $^{- 1}$ is

1

52.4 \times 10^{- 4}

2

524 \times 10^{- 4}

3

402 \times 10^{- 4}

4

262 \times 10^{- 4}

Explanation:

$\begin{aligned} Given that, \\ {BaCl}_{2} = 280 \times 10^{- 4} \\ NaOH = 248 \times 10^{- 4} \\ NaCl = 126 \times 10^{- 4} \\ {BaCl}_{2} + 2 NaOH ⟶ Ba (OH)_{2} + 2 NaCl \\ λ_{mBa (OH)_{2}}^{\infty} = λ_{mBaCl}^{2} + 2 λ_{m NaOH}^{\infty} - 2 λ_{m NaCl}^{\infty} \\ = 280 \times 10^{- 4} + 2 \times 248 \times 10^{- 4} - 2 \times 126 \times 10^{- 4} \\ = (280 + 496 - 252) \times 10^{- 4} \\ = 524 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \end{aligned}$

ELECTROCHEMISTRY

276135 The molar conductance of $NaCl, HCl$ and ${CH}_{3} COONa$ at infinite dilution are 126.45, 426.16 and $91.0 S {cm}^{2} {mol}^{- 1}$ respectively. The molar conductance of ${CH}_{3} COOH$ at infinte dilution is Choose the right option for your answer.

1

540.48 S {cm}^{2} {mol}^{- 1}

2

201.28 {Scm}^{2} {mol}^{- 1}

3

390.71 S {cm}^{2} {mol}^{- 1}

4

698.28 {Scm}^{2} {mol}^{- 1}

Explanation:

Given that,
$\begin{aligned} Λ_{m}^{\circ} NaCl = 126.5 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} HCl = 426.16 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} {CH}_{3} COONa = 91.0 S {cm}^{2} {mol}^{- 1} \end{aligned}$
$Λ_{m}^{\circ} {CH}_{3} COOH = Λ_{m}^{\circ} {CH}_{3} COONa + Λ_{m}^{\circ} HCl - Λ_{m}^{\circ} NaCl$
$\begin{aligned} = 91.0 + 426.16 - 126.5 \\ = 390.71 S {cm}^{2} {mol}^{- 1} \end{aligned}$

MHT CET-03.05.2019

ELECTROCHEMISTRY

276136 Unit of equivalent conductance is :

1

{ohm}^{- 1} {cm}^{2} (g - eq)^{- 1}

2

ohm cm (g - eq)

3

ohm cm^{2} (g - eq)^{- 1}

4

{ohm}^{- 1} cm (g - eq)^{- 1}

SRM JEE 2014

Explanation:

The equivalent conductance is calculated by multiplying the specific conductivity and the volume of the solution having one gram equivalent of the electrolyte.
$\begin{aligned} Equivalent conductivity = κ \times \frac{1000}{Normality} \\ = \frac{{ohm}^{- 1} cm}{g_{equivalent {cm}^{- 1}}} \\ {= {ohm}^{- 1} {cm}^{2} (g equivalent)}^{- 1} \\ = {ohm}^{- 1} {cm}^{2} (g - eq)^{- 1} \end{aligned}$

2006

ELECTROCHEMISTRY

276138 Equivalent conductance's of ${Ba}^{2 +}$ and ${Cl}^{-}$ ions are 127 and $76 {ohm}^{- 1} {cm}^{- 1} {eq}^{- 1}$ respectively. Equivalent conductance of ${BaCl}_{2}$ at infinite dilution is

1 139.5

2 101.5

3 203

4 279

Explanation:

Chemical equation-
${BaCl}_{2} \to {Ba}^{2 +} + 2 {Cl}^{-}$
From the Kohlrausch law -*
$\begin{aligned} λ_{\infty} = \frac{1}{n} λ_{+}^{\infty} + \frac{1}{n} λ_{-}^{\infty} \\ λ_{\infty} ({BaCl}_{2}) = \frac{1}{2} \times λ_{{Ba}^{2 +}}^{\infty} + \frac{1}{1} \times λ_{{Cl}^{-}}^{\infty} \\ = \frac{127}{2} + 76 \\ = 63.5 + 76 \\ = 139.5 {ohm}^{- 1} {cm}^{- 1} {eq}^{- 1} \end{aligned}$

ELECTROCHEMISTRY

276133 Resistance of a conductivity cell filled with 0.1 mol $L^{- 1} NaCl$ is $100 Ohm$ . If the resistance of the same cell when filled with $0.02 mol L^{- 1} NaCl$ solution is $258 Ohm$ . The conductivity of $0.02 mol L^{- 1} NaCl$ solution is (Conductivity of $0.1 L^{- 1} NaCl$ is $1.29 {Sm}^{- 1}$ )

1

1.0 S m^{- 1}

2

0.2 S m^{- 1}

3

2.0 S m^{- 1}

4

0.5 S m^{- 1}

Explanation:

For $0.1 mol L^{- 1} NaCl$ solution,
Conductivity $(K) = 1.29 S m^{- 1}$ , Resistance $(R) = 100 Ω$
$∵$ Cell constant $=$ Conductivity $\times$ Resistance
$= 1.29 \times 100 = 129$
For, $0.02 mol L^{- 1} NaCl$ solution
Resistance $= 258 Ω$
$\begin{aligned} Conductivity (K) & = \frac{Cell constant}{Resistance} \\ = \frac{129}{258} \\ = 0.5 S m^{- 1} \end{aligned}$

AP-EAMCET-05.07.2022

ELECTROCHEMISTRY

276134 At $25^{\circ} C$ , the molar conductances at infinite dilution for the strong electrolytes $NaOH, NaCl$ and ${BaCl}_{2}$ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times$ $10^{- 4} {Sm}^{2} {mol}^{- 1}$ respectively, $λ_{m}^{\circ} Ba (OH)_{2}$ in ${Sm}^{2}$ mol $^{- 1}$ is

1

52.4 \times 10^{- 4}

2

524 \times 10^{- 4}

3

402 \times 10^{- 4}

4

262 \times 10^{- 4}

Explanation:

$\begin{aligned} Given that, \\ {BaCl}_{2} = 280 \times 10^{- 4} \\ NaOH = 248 \times 10^{- 4} \\ NaCl = 126 \times 10^{- 4} \\ {BaCl}_{2} + 2 NaOH ⟶ Ba (OH)_{2} + 2 NaCl \\ λ_{mBa (OH)_{2}}^{\infty} = λ_{mBaCl}^{2} + 2 λ_{m NaOH}^{\infty} - 2 λ_{m NaCl}^{\infty} \\ = 280 \times 10^{- 4} + 2 \times 248 \times 10^{- 4} - 2 \times 126 \times 10^{- 4} \\ = (280 + 496 - 252) \times 10^{- 4} \\ = 524 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \end{aligned}$

ELECTROCHEMISTRY

276135 The molar conductance of $NaCl, HCl$ and ${CH}_{3} COONa$ at infinite dilution are 126.45, 426.16 and $91.0 S {cm}^{2} {mol}^{- 1}$ respectively. The molar conductance of ${CH}_{3} COOH$ at infinte dilution is Choose the right option for your answer.

1

540.48 S {cm}^{2} {mol}^{- 1}

2

201.28 {Scm}^{2} {mol}^{- 1}

3

390.71 S {cm}^{2} {mol}^{- 1}

4

698.28 {Scm}^{2} {mol}^{- 1}

Explanation:

Given that,
$\begin{aligned} Λ_{m}^{\circ} NaCl = 126.5 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} HCl = 426.16 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} {CH}_{3} COONa = 91.0 S {cm}^{2} {mol}^{- 1} \end{aligned}$
$Λ_{m}^{\circ} {CH}_{3} COOH = Λ_{m}^{\circ} {CH}_{3} COONa + Λ_{m}^{\circ} HCl - Λ_{m}^{\circ} NaCl$
$\begin{aligned} = 91.0 + 426.16 - 126.5 \\ = 390.71 S {cm}^{2} {mol}^{- 1} \end{aligned}$

MHT CET-03.05.2019