QBManager

ELECTROCHEMISTRY

276219 Which of the following expressions correctly represents the equivalent conductance at infinite dilution of ${Al}_{2} {({SO}_{4})}_{3}$ ?. Given that $Λ_{{Al}^{3 +}}^{\circ}$ and $Λ_{{SO}_{4}^{2 -}}^{\circ}$ are the equivalent conductances at infinite dilution of the respective ions.

1

2 Λ_{{Al}^{3 +}}^{\circ} + 3 Λ_{{SO}_{4}^{2 -}}^{\circ}

2

Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}

3

(Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}) \times 6

4

\frac{1}{3} Λ_{{Al}^{3 +}}^{\circ} + \frac{1}{2} Λ_{{SO}_{4}^{2 -}}^{\circ}

Explanation:

${Al}_{2} {({SO}_{4})}_{3} ⇌ 2 {Al}^{3 +} + 3 {SO}_{4}^{2 -}$
At infinite dilution dissociation of electrolytic solution is complete and each ion makes a definite contribution toward molar conductance of each electrolyte.
We can calculate the equivalent conductance only ions, so the equivalent conductance at infinite dilution.
$\land_{{Al}_{2} {({SO}_{4})}_{3}}^{\circ} = Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{0}$
$Λ_{eq}^{\circ} = Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}$

ELECTROCHEMISTRY

276222 A conductivity cell has been calibrated with a 0.01M 1 : 1 electrolyte solution (specific conductance, $k = 1.25 \times 10^{- 3} S {cm}^{- 1}$ ) in the cell and the measured resistance was $800 Ω$ at $25^{\circ} C$ . The cell constant will be

1

1.02 {cm}^{- 1}

2

0.102 {cm}^{- 1}

3

1.00 {cm}^{- 1}

4

0.5 {cm}^{- 1}

Explanation:

Given that $κ = 1.25 \times 10^{- 3} s / cm$
$R = 800 Ω$ We know,
$\begin{aligned} R = ρ \frac{l}{A} and ρ = \frac{1}{κ} \\ 800 = \frac{1}{1.25 \times 10^{- 3}} \times (\frac{l}{A}) \\ (\frac{l}{A} = cell constant) \\ \frac{l}{A} = 800 \times 1.25 \times 10^{- 3} \\ = 1 {cm}^{- 1} \end{aligned}$

WB-JEE-2013

ELECTROCHEMISTRY

276224 At $25^{\circ} C$ , the molar conductance of $0.007 M$ hydrofluoric acid is $150 mho {cm}^{2} {mol}^{- 1}$ and it's $A^{0} m = 500 mho {cm}^{2} {mol}^{- 1}$ The value of the dissociation constant of the acid at the given concentration at $25^{\circ} C$ is

1

7 \times 10^{- 4} M

2

7 \times 10^{- 5} M

3

9 \times 10^{- 3} M

4

9 \times 10^{- 4} M

Explanation:

Given that Molar conductance $(C) = 0.007 M$ $α = \frac{λ_{m}}{λ_{m}^{o}}$
$\begin{aligned} α & = \frac{150}{500} = 0.3 \\ HF & ⇌ H^{+} + F^{-} \\ K & = C \frac{α^{2}}{1 - α} = \frac{0.007 \times (0.3)^{2}}{1 - 0.3} \\ = \frac{0.007 \times 0.09}{0.7} \\ = 9 \times 10^{- 4} M . \end{aligned}$

WB-JEE-2014

ELECTROCHEMISTRY

276226 Molar conductances of ${BaCl}_{2}, H_{2} {SO}_{4}$ and $HCl$ at infinite dilutions are $x_{1}, x_{2}$ and $x_{3}$ respectively. Equivalent conductance of ${BaSO}_{4}$ at infinite dilution will be:

1

(x_{1} + x_{2} - x_{3}) / 2

2

x_{1} + x_{2} - 2 x_{3}

3

(x_{1} - x_{2} - x_{3}) / 2

4

(x_{1} + x_{2} - 2 x_{3}) / 2

Explanation:

$Λ_{m} \cdot {BaCl}_{2} = Λ_{{mBe}^{2 +}} + 2 Λ_{{mCl}^{-}} x_{1}$
$Λ_{m} \cdot H_{2} {SO}_{4} = 2 Λ_{{mH}^{+}} + Λ_{{mSO}_{4}^{-}} x_{2}$
$Λ_{mHCl} = Λ_{{mH}^{+}} + Λ_{{mCl}^{-}} x_{3}] \times 2$
$∴ Λ_{m} {BaSO}_{4} = (x_{1} + x_{2} - 2 x_{3})$
$Λ_{eq} {BaSO}_{4} = \frac{Λ_{m} \cdot {BaSO}_{4}}{Total charge (cation / anion)}$
$Λ_{eq} {BaSO}_{4} = \frac{[x_{1} + x_{2} - 2 x_{3}]}{2}$

AIIMS-2011

ELECTROCHEMISTRY

276219 Which of the following expressions correctly represents the equivalent conductance at infinite dilution of ${Al}_{2} {({SO}_{4})}_{3}$ ?. Given that $Λ_{{Al}^{3 +}}^{\circ}$ and $Λ_{{SO}_{4}^{2 -}}^{\circ}$ are the equivalent conductances at infinite dilution of the respective ions.

1

2 Λ_{{Al}^{3 +}}^{\circ} + 3 Λ_{{SO}_{4}^{2 -}}^{\circ}

2

Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}

3

(Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}) \times 6

4

\frac{1}{3} Λ_{{Al}^{3 +}}^{\circ} + \frac{1}{2} Λ_{{SO}_{4}^{2 -}}^{\circ}

Explanation:

${Al}_{2} {({SO}_{4})}_{3} ⇌ 2 {Al}^{3 +} + 3 {SO}_{4}^{2 -}$
At infinite dilution dissociation of electrolytic solution is complete and each ion makes a definite contribution toward molar conductance of each electrolyte.
We can calculate the equivalent conductance only ions, so the equivalent conductance at infinite dilution.
$\land_{{Al}_{2} {({SO}_{4})}_{3}}^{\circ} = Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{0}$
$Λ_{eq}^{\circ} = Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}$

ELECTROCHEMISTRY

276222 A conductivity cell has been calibrated with a 0.01M 1 : 1 electrolyte solution (specific conductance, $k = 1.25 \times 10^{- 3} S {cm}^{- 1}$ ) in the cell and the measured resistance was $800 Ω$ at $25^{\circ} C$ . The cell constant will be

1

1.02 {cm}^{- 1}

2

0.102 {cm}^{- 1}

3

1.00 {cm}^{- 1}

4

0.5 {cm}^{- 1}

Explanation:

Given that $κ = 1.25 \times 10^{- 3} s / cm$
$R = 800 Ω$ We know,
$\begin{aligned} R = ρ \frac{l}{A} and ρ = \frac{1}{κ} \\ 800 = \frac{1}{1.25 \times 10^{- 3}} \times (\frac{l}{A}) \\ (\frac{l}{A} = cell constant) \\ \frac{l}{A} = 800 \times 1.25 \times 10^{- 3} \\ = 1 {cm}^{- 1} \end{aligned}$

WB-JEE-2013

ELECTROCHEMISTRY

276224 At $25^{\circ} C$ , the molar conductance of $0.007 M$ hydrofluoric acid is $150 mho {cm}^{2} {mol}^{- 1}$ and it's $A^{0} m = 500 mho {cm}^{2} {mol}^{- 1}$ The value of the dissociation constant of the acid at the given concentration at $25^{\circ} C$ is

1

7 \times 10^{- 4} M

2

7 \times 10^{- 5} M

3

9 \times 10^{- 3} M

4

9 \times 10^{- 4} M

Explanation:

Given that Molar conductance $(C) = 0.007 M$ $α = \frac{λ_{m}}{λ_{m}^{o}}$
$\begin{aligned} α & = \frac{150}{500} = 0.3 \\ HF & ⇌ H^{+} + F^{-} \\ K & = C \frac{α^{2}}{1 - α} = \frac{0.007 \times (0.3)^{2}}{1 - 0.3} \\ = \frac{0.007 \times 0.09}{0.7} \\ = 9 \times 10^{- 4} M . \end{aligned}$

WB-JEE-2014

ELECTROCHEMISTRY

276226 Molar conductances of ${BaCl}_{2}, H_{2} {SO}_{4}$ and $HCl$ at infinite dilutions are $x_{1}, x_{2}$ and $x_{3}$ respectively. Equivalent conductance of ${BaSO}_{4}$ at infinite dilution will be:

1

(x_{1} + x_{2} - x_{3}) / 2

2

x_{1} + x_{2} - 2 x_{3}

3

(x_{1} - x_{2} - x_{3}) / 2

4

(x_{1} + x_{2} - 2 x_{3}) / 2

Explanation:

$Λ_{m} \cdot {BaCl}_{2} = Λ_{{mBe}^{2 +}} + 2 Λ_{{mCl}^{-}} x_{1}$
$Λ_{m} \cdot H_{2} {SO}_{4} = 2 Λ_{{mH}^{+}} + Λ_{{mSO}_{4}^{-}} x_{2}$
$Λ_{mHCl} = Λ_{{mH}^{+}} + Λ_{{mCl}^{-}} x_{3}] \times 2$
$∴ Λ_{m} {BaSO}_{4} = (x_{1} + x_{2} - 2 x_{3})$
$Λ_{eq} {BaSO}_{4} = \frac{Λ_{m} \cdot {BaSO}_{4}}{Total charge (cation / anion)}$
$Λ_{eq} {BaSO}_{4} = \frac{[x_{1} + x_{2} - 2 x_{3}]}{2}$

AIIMS-2011

ELECTROCHEMISTRY

276219 Which of the following expressions correctly represents the equivalent conductance at infinite dilution of ${Al}_{2} {({SO}_{4})}_{3}$ ?. Given that $Λ_{{Al}^{3 +}}^{\circ}$ and $Λ_{{SO}_{4}^{2 -}}^{\circ}$ are the equivalent conductances at infinite dilution of the respective ions.

1

2 Λ_{{Al}^{3 +}}^{\circ} + 3 Λ_{{SO}_{4}^{2 -}}^{\circ}

2

Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}

3

(Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}) \times 6

4

\frac{1}{3} Λ_{{Al}^{3 +}}^{\circ} + \frac{1}{2} Λ_{{SO}_{4}^{2 -}}^{\circ}

Explanation:

${Al}_{2} {({SO}_{4})}_{3} ⇌ 2 {Al}^{3 +} + 3 {SO}_{4}^{2 -}$
At infinite dilution dissociation of electrolytic solution is complete and each ion makes a definite contribution toward molar conductance of each electrolyte.
We can calculate the equivalent conductance only ions, so the equivalent conductance at infinite dilution.
$\land_{{Al}_{2} {({SO}_{4})}_{3}}^{\circ} = Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{0}$
$Λ_{eq}^{\circ} = Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}$

ELECTROCHEMISTRY

276222 A conductivity cell has been calibrated with a 0.01M 1 : 1 electrolyte solution (specific conductance, $k = 1.25 \times 10^{- 3} S {cm}^{- 1}$ ) in the cell and the measured resistance was $800 Ω$ at $25^{\circ} C$ . The cell constant will be

1

1.02 {cm}^{- 1}

2

0.102 {cm}^{- 1}

3

1.00 {cm}^{- 1}

4

0.5 {cm}^{- 1}

Explanation:

Given that $κ = 1.25 \times 10^{- 3} s / cm$
$R = 800 Ω$ We know,
$\begin{aligned} R = ρ \frac{l}{A} and ρ = \frac{1}{κ} \\ 800 = \frac{1}{1.25 \times 10^{- 3}} \times (\frac{l}{A}) \\ (\frac{l}{A} = cell constant) \\ \frac{l}{A} = 800 \times 1.25 \times 10^{- 3} \\ = 1 {cm}^{- 1} \end{aligned}$

WB-JEE-2013

ELECTROCHEMISTRY

276224 At $25^{\circ} C$ , the molar conductance of $0.007 M$ hydrofluoric acid is $150 mho {cm}^{2} {mol}^{- 1}$ and it's $A^{0} m = 500 mho {cm}^{2} {mol}^{- 1}$ The value of the dissociation constant of the acid at the given concentration at $25^{\circ} C$ is

1

7 \times 10^{- 4} M

2

7 \times 10^{- 5} M

3

9 \times 10^{- 3} M

4

9 \times 10^{- 4} M

Explanation:

Given that Molar conductance $(C) = 0.007 M$ $α = \frac{λ_{m}}{λ_{m}^{o}}$
$\begin{aligned} α & = \frac{150}{500} = 0.3 \\ HF & ⇌ H^{+} + F^{-} \\ K & = C \frac{α^{2}}{1 - α} = \frac{0.007 \times (0.3)^{2}}{1 - 0.3} \\ = \frac{0.007 \times 0.09}{0.7} \\ = 9 \times 10^{- 4} M . \end{aligned}$

WB-JEE-2014

ELECTROCHEMISTRY

276226 Molar conductances of ${BaCl}_{2}, H_{2} {SO}_{4}$ and $HCl$ at infinite dilutions are $x_{1}, x_{2}$ and $x_{3}$ respectively. Equivalent conductance of ${BaSO}_{4}$ at infinite dilution will be:

1

(x_{1} + x_{2} - x_{3}) / 2

2

x_{1} + x_{2} - 2 x_{3}

3

(x_{1} - x_{2} - x_{3}) / 2

4

(x_{1} + x_{2} - 2 x_{3}) / 2

Explanation:

$Λ_{m} \cdot {BaCl}_{2} = Λ_{{mBe}^{2 +}} + 2 Λ_{{mCl}^{-}} x_{1}$
$Λ_{m} \cdot H_{2} {SO}_{4} = 2 Λ_{{mH}^{+}} + Λ_{{mSO}_{4}^{-}} x_{2}$
$Λ_{mHCl} = Λ_{{mH}^{+}} + Λ_{{mCl}^{-}} x_{3}] \times 2$
$∴ Λ_{m} {BaSO}_{4} = (x_{1} + x_{2} - 2 x_{3})$
$Λ_{eq} {BaSO}_{4} = \frac{Λ_{m} \cdot {BaSO}_{4}}{Total charge (cation / anion)}$
$Λ_{eq} {BaSO}_{4} = \frac{[x_{1} + x_{2} - 2 x_{3}]}{2}$

AIIMS-2011

ELECTROCHEMISTRY

276219 Which of the following expressions correctly represents the equivalent conductance at infinite dilution of ${Al}_{2} {({SO}_{4})}_{3}$ ?. Given that $Λ_{{Al}^{3 +}}^{\circ}$ and $Λ_{{SO}_{4}^{2 -}}^{\circ}$ are the equivalent conductances at infinite dilution of the respective ions.

1

2 Λ_{{Al}^{3 +}}^{\circ} + 3 Λ_{{SO}_{4}^{2 -}}^{\circ}

2

Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}

3

(Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}) \times 6

4

\frac{1}{3} Λ_{{Al}^{3 +}}^{\circ} + \frac{1}{2} Λ_{{SO}_{4}^{2 -}}^{\circ}

Explanation:

${Al}_{2} {({SO}_{4})}_{3} ⇌ 2 {Al}^{3 +} + 3 {SO}_{4}^{2 -}$
At infinite dilution dissociation of electrolytic solution is complete and each ion makes a definite contribution toward molar conductance of each electrolyte.
We can calculate the equivalent conductance only ions, so the equivalent conductance at infinite dilution.
$\land_{{Al}_{2} {({SO}_{4})}_{3}}^{\circ} = Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{0}$
$Λ_{eq}^{\circ} = Λ_{{Al}^{3 +}}^{\circ} + Λ_{{SO}_{4}^{2 -}}^{\circ}$

ELECTROCHEMISTRY

276222 A conductivity cell has been calibrated with a 0.01M 1 : 1 electrolyte solution (specific conductance, $k = 1.25 \times 10^{- 3} S {cm}^{- 1}$ ) in the cell and the measured resistance was $800 Ω$ at $25^{\circ} C$ . The cell constant will be

1

1.02 {cm}^{- 1}

2

0.102 {cm}^{- 1}

3

1.00 {cm}^{- 1}

4

0.5 {cm}^{- 1}

Explanation:

Given that $κ = 1.25 \times 10^{- 3} s / cm$
$R = 800 Ω$ We know,
$\begin{aligned} R = ρ \frac{l}{A} and ρ = \frac{1}{κ} \\ 800 = \frac{1}{1.25 \times 10^{- 3}} \times (\frac{l}{A}) \\ (\frac{l}{A} = cell constant) \\ \frac{l}{A} = 800 \times 1.25 \times 10^{- 3} \\ = 1 {cm}^{- 1} \end{aligned}$

WB-JEE-2013

ELECTROCHEMISTRY

276224 At $25^{\circ} C$ , the molar conductance of $0.007 M$ hydrofluoric acid is $150 mho {cm}^{2} {mol}^{- 1}$ and it's $A^{0} m = 500 mho {cm}^{2} {mol}^{- 1}$ The value of the dissociation constant of the acid at the given concentration at $25^{\circ} C$ is

1

7 \times 10^{- 4} M

2

7 \times 10^{- 5} M

3

9 \times 10^{- 3} M

4

9 \times 10^{- 4} M

Explanation:

Given that Molar conductance $(C) = 0.007 M$ $α = \frac{λ_{m}}{λ_{m}^{o}}$
$\begin{aligned} α & = \frac{150}{500} = 0.3 \\ HF & ⇌ H^{+} + F^{-} \\ K & = C \frac{α^{2}}{1 - α} = \frac{0.007 \times (0.3)^{2}}{1 - 0.3} \\ = \frac{0.007 \times 0.09}{0.7} \\ = 9 \times 10^{- 4} M . \end{aligned}$

WB-JEE-2014

ELECTROCHEMISTRY

276226 Molar conductances of ${BaCl}_{2}, H_{2} {SO}_{4}$ and $HCl$ at infinite dilutions are $x_{1}, x_{2}$ and $x_{3}$ respectively. Equivalent conductance of ${BaSO}_{4}$ at infinite dilution will be:

1

(x_{1} + x_{2} - x_{3}) / 2

2

x_{1} + x_{2} - 2 x_{3}

3

(x_{1} - x_{2} - x_{3}) / 2

4

(x_{1} + x_{2} - 2 x_{3}) / 2

Explanation:

$Λ_{m} \cdot {BaCl}_{2} = Λ_{{mBe}^{2 +}} + 2 Λ_{{mCl}^{-}} x_{1}$
$Λ_{m} \cdot H_{2} {SO}_{4} = 2 Λ_{{mH}^{+}} + Λ_{{mSO}_{4}^{-}} x_{2}$
$Λ_{mHCl} = Λ_{{mH}^{+}} + Λ_{{mCl}^{-}} x_{3}] \times 2$
$∴ Λ_{m} {BaSO}_{4} = (x_{1} + x_{2} - 2 x_{3})$
$Λ_{eq} {BaSO}_{4} = \frac{Λ_{m} \cdot {BaSO}_{4}}{Total charge (cation / anion)}$
$Λ_{eq} {BaSO}_{4} = \frac{[x_{1} + x_{2} - 2 x_{3}]}{2}$

AIIMS-2011