QBManager

CHXII03:ELECTROCHEMISTRY

330328 If molar conductivity of ${Ca}^{2 +}$ and ${Cl}^{-}$ ions are 119 and $71 S {cm}^{2} {mol}^{- 1}$ respectively, then the molar conductivity of ${CaCl}_{2}$ at infinite dilution is

1

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

2

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

3

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

4

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

Explanation:

Given, $Λ_{C a^{2 +}}^{\circ} = 119 S c m^{2} mo l^{- 1}$
$Λ_{{Cl}^{-}}^{\circ} = 71 S {cm}^{2} {mol}^{- 1}$
As we know that,
$Λ_{{CaCl}_{2}}^{\circ} = Λ_{{Ca}^{2 +}}^{\circ} + 2 \times Λ_{{Cl}^{-}}^{\circ}$
$∴ Λ_{C a C l_{2}}^{\circ} = 119 + 2 \times 71$
$Λ_{C a C l_{2}}^{\circ} = 261 S c m^{2} m o l^{- 1}$

CHXII03:ELECTROCHEMISTRY

330329 According to Kohlrausch law, the limiting value of molar conductance of an electrolyte $A_{2} B$ is

1

λ_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}

2

{2 λ}_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}

3

λ_{A^{+}}^{^{\circ}} - λ_{B^{- 2}}^{^{\circ}}

4

λ_{A^{+}}^{^{\circ}} + \frac{1}{2} λ_{B^{- 2}}^{^{\circ}}

Explanation:

$Λ_{A_{2} B}^{^{\circ}} {= 2 λ}_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}$

CHXII03:ELECTROCHEMISTRY

330330 At Infinite dilution, each ion makes definite contribution to conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte. This law states that

1 Kohlrausch’s law

2 Debye Huckel rule

3 Faraday’s law

4 Arrhenius law

CHXII03:ELECTROCHEMISTRY

330331 The resistance of $0.1 M$ weak acid HA in a conductivity cell is $2 \times 10^{3} Ohm$ . The cell constant of the cell is $0.78 {cm}^{- 1}$ and $Λ_{m}^{\circ}$ of acid HA is $390 {Scm}^{2} {mol}^{- 1}$ . The $pH$ of the solution is

1 5

2 3

3 3.3

4 4.2

Explanation:

$C = 0.1 M; Λ_{m}^{0} = 390 {Scm}^{2} {mol}^{- 1}$
$\begin{aligned} R = 2 \times 10^{3} ohm \\ G^{*} = 0.78 {cm}^{- 1} \\ K = \frac{R}{G^{*}} = \frac{0.78}{2 \times 10^{3}} = 3.9 \times 10^{- 4} \\ Λ_{m} = \frac{K \times 1000}{C} = \frac{3.9 \times 10^{- 4} \times 1000}{0.1} = 3.9 \\ α = \frac{Λ_{m}}{Λ_{C}^{0}} = \frac{3.9}{390} = 10^{- 2} \\ [H^{+}] = c α = 0.1 \times 10^{- 2} = 10^{- 3} \\ pH = - \log 10^{- 3} = 3 \end{aligned}$

KCET - 2023

CHXII03:ELECTROCHEMISTRY

330333 The molar conductivity of 0.007 M acetic acid is $20 S . c m^{2} m o l^{- 1}$ .
What is the dissociation constant of acetic acid ? Choose the correct option.
$[\begin{array}{l} Λ_{H^{+}}^{o} = 350 S c m^{2} m o l^{- 1}; \\ Λ_{C H_{3} C O O^{-}}^{o} = 50 S c m^{2} m o l^{- 1} \end{array}]$

1

2 .50 \times 1 0^{- 4} m o l L^{- 1}

2

1 .75 \times 1 0^{- 5} m o l L^{- 1}

3

2 .50 \times 1 0^{- 5} m o l L^{- 1}

4

1 .75 \times 1 0^{- 4} m o l L^{- 1}

Explanation:

$α = \frac{{[Λ]}_{M}^{C}}{{[Λ]}_{M}^{\infty}} = \frac{{[Λ]}_{M}^{C}}{{[Λ]}_{(H^{+})}^{\infty} {[Λ]}_{C H_{3} C O O^{-}}^{\infty}} = \frac{20}{350 + 50}$
$= \frac{20}{400} = \frac{1}{20}$
$K_{a} = \frac{C α^{2}}{1 - α}$
$K_{a} = C α^{2}$
$= 0 .007 \times {(\frac{1}{20})}^{2} = \frac{7 \times 1 0^{- 3}}{400}$
$K_{a} = \frac{7}{4} \times 1 0^{- 5}$
$= 1 .75 \times 1 0^{- 5} m o l L^{- 1}$

NEET - 2021

CHXII03:ELECTROCHEMISTRY

330328 If molar conductivity of ${Ca}^{2 +}$ and ${Cl}^{-}$ ions are 119 and $71 S {cm}^{2} {mol}^{- 1}$ respectively, then the molar conductivity of ${CaCl}_{2}$ at infinite dilution is

1

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

2

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

3

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

4

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

Explanation:

Given, $Λ_{C a^{2 +}}^{\circ} = 119 S c m^{2} mo l^{- 1}$
$Λ_{{Cl}^{-}}^{\circ} = 71 S {cm}^{2} {mol}^{- 1}$
As we know that,
$Λ_{{CaCl}_{2}}^{\circ} = Λ_{{Ca}^{2 +}}^{\circ} + 2 \times Λ_{{Cl}^{-}}^{\circ}$
$∴ Λ_{C a C l_{2}}^{\circ} = 119 + 2 \times 71$
$Λ_{C a C l_{2}}^{\circ} = 261 S c m^{2} m o l^{- 1}$

CHXII03:ELECTROCHEMISTRY

330329 According to Kohlrausch law, the limiting value of molar conductance of an electrolyte $A_{2} B$ is

1

λ_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}

2

{2 λ}_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}

3

λ_{A^{+}}^{^{\circ}} - λ_{B^{- 2}}^{^{\circ}}

4

λ_{A^{+}}^{^{\circ}} + \frac{1}{2} λ_{B^{- 2}}^{^{\circ}}

Explanation:

$Λ_{A_{2} B}^{^{\circ}} {= 2 λ}_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}$

CHXII03:ELECTROCHEMISTRY

330330 At Infinite dilution, each ion makes definite contribution to conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte. This law states that

1 Kohlrausch’s law

2 Debye Huckel rule

3 Faraday’s law

4 Arrhenius law

CHXII03:ELECTROCHEMISTRY

330331 The resistance of $0.1 M$ weak acid HA in a conductivity cell is $2 \times 10^{3} Ohm$ . The cell constant of the cell is $0.78 {cm}^{- 1}$ and $Λ_{m}^{\circ}$ of acid HA is $390 {Scm}^{2} {mol}^{- 1}$ . The $pH$ of the solution is

1 5

2 3

3 3.3

4 4.2

Explanation:

$C = 0.1 M; Λ_{m}^{0} = 390 {Scm}^{2} {mol}^{- 1}$
$\begin{aligned} R = 2 \times 10^{3} ohm \\ G^{*} = 0.78 {cm}^{- 1} \\ K = \frac{R}{G^{*}} = \frac{0.78}{2 \times 10^{3}} = 3.9 \times 10^{- 4} \\ Λ_{m} = \frac{K \times 1000}{C} = \frac{3.9 \times 10^{- 4} \times 1000}{0.1} = 3.9 \\ α = \frac{Λ_{m}}{Λ_{C}^{0}} = \frac{3.9}{390} = 10^{- 2} \\ [H^{+}] = c α = 0.1 \times 10^{- 2} = 10^{- 3} \\ pH = - \log 10^{- 3} = 3 \end{aligned}$

KCET - 2023

CHXII03:ELECTROCHEMISTRY

330333 The molar conductivity of 0.007 M acetic acid is $20 S . c m^{2} m o l^{- 1}$ .
What is the dissociation constant of acetic acid ? Choose the correct option.
$[\begin{array}{l} Λ_{H^{+}}^{o} = 350 S c m^{2} m o l^{- 1}; \\ Λ_{C H_{3} C O O^{-}}^{o} = 50 S c m^{2} m o l^{- 1} \end{array}]$

1

2 .50 \times 1 0^{- 4} m o l L^{- 1}

2

1 .75 \times 1 0^{- 5} m o l L^{- 1}

3

2 .50 \times 1 0^{- 5} m o l L^{- 1}

4

1 .75 \times 1 0^{- 4} m o l L^{- 1}

Explanation:

$α = \frac{{[Λ]}_{M}^{C}}{{[Λ]}_{M}^{\infty}} = \frac{{[Λ]}_{M}^{C}}{{[Λ]}_{(H^{+})}^{\infty} {[Λ]}_{C H_{3} C O O^{-}}^{\infty}} = \frac{20}{350 + 50}$
$= \frac{20}{400} = \frac{1}{20}$
$K_{a} = \frac{C α^{2}}{1 - α}$
$K_{a} = C α^{2}$
$= 0 .007 \times {(\frac{1}{20})}^{2} = \frac{7 \times 1 0^{- 3}}{400}$
$K_{a} = \frac{7}{4} \times 1 0^{- 5}$
$= 1 .75 \times 1 0^{- 5} m o l L^{- 1}$

NEET - 2021

CHXII03:ELECTROCHEMISTRY

330328 If molar conductivity of ${Ca}^{2 +}$ and ${Cl}^{-}$ ions are 119 and $71 S {cm}^{2} {mol}^{- 1}$ respectively, then the molar conductivity of ${CaCl}_{2}$ at infinite dilution is

1

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

2

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

3

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

4

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

Explanation:

Given, $Λ_{C a^{2 +}}^{\circ} = 119 S c m^{2} mo l^{- 1}$
$Λ_{{Cl}^{-}}^{\circ} = 71 S {cm}^{2} {mol}^{- 1}$
As we know that,
$Λ_{{CaCl}_{2}}^{\circ} = Λ_{{Ca}^{2 +}}^{\circ} + 2 \times Λ_{{Cl}^{-}}^{\circ}$
$∴ Λ_{C a C l_{2}}^{\circ} = 119 + 2 \times 71$
$Λ_{C a C l_{2}}^{\circ} = 261 S c m^{2} m o l^{- 1}$

CHXII03:ELECTROCHEMISTRY

330329 According to Kohlrausch law, the limiting value of molar conductance of an electrolyte $A_{2} B$ is

1

λ_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}

2

{2 λ}_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}

3

λ_{A^{+}}^{^{\circ}} - λ_{B^{- 2}}^{^{\circ}}

4

λ_{A^{+}}^{^{\circ}} + \frac{1}{2} λ_{B^{- 2}}^{^{\circ}}

Explanation:

$Λ_{A_{2} B}^{^{\circ}} {= 2 λ}_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}$

CHXII03:ELECTROCHEMISTRY

330330 At Infinite dilution, each ion makes definite contribution to conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte. This law states that

1 Kohlrausch’s law

2 Debye Huckel rule

3 Faraday’s law

4 Arrhenius law

CHXII03:ELECTROCHEMISTRY

330331 The resistance of $0.1 M$ weak acid HA in a conductivity cell is $2 \times 10^{3} Ohm$ . The cell constant of the cell is $0.78 {cm}^{- 1}$ and $Λ_{m}^{\circ}$ of acid HA is $390 {Scm}^{2} {mol}^{- 1}$ . The $pH$ of the solution is

1 5

2 3

3 3.3

4 4.2

Explanation:

$C = 0.1 M; Λ_{m}^{0} = 390 {Scm}^{2} {mol}^{- 1}$
$\begin{aligned} R = 2 \times 10^{3} ohm \\ G^{*} = 0.78 {cm}^{- 1} \\ K = \frac{R}{G^{*}} = \frac{0.78}{2 \times 10^{3}} = 3.9 \times 10^{- 4} \\ Λ_{m} = \frac{K \times 1000}{C} = \frac{3.9 \times 10^{- 4} \times 1000}{0.1} = 3.9 \\ α = \frac{Λ_{m}}{Λ_{C}^{0}} = \frac{3.9}{390} = 10^{- 2} \\ [H^{+}] = c α = 0.1 \times 10^{- 2} = 10^{- 3} \\ pH = - \log 10^{- 3} = 3 \end{aligned}$

KCET - 2023

CHXII03:ELECTROCHEMISTRY

330333 The molar conductivity of 0.007 M acetic acid is $20 S . c m^{2} m o l^{- 1}$ .
What is the dissociation constant of acetic acid ? Choose the correct option.
$[\begin{array}{l} Λ_{H^{+}}^{o} = 350 S c m^{2} m o l^{- 1}; \\ Λ_{C H_{3} C O O^{-}}^{o} = 50 S c m^{2} m o l^{- 1} \end{array}]$

1

2 .50 \times 1 0^{- 4} m o l L^{- 1}

2

1 .75 \times 1 0^{- 5} m o l L^{- 1}

3

2 .50 \times 1 0^{- 5} m o l L^{- 1}

4

1 .75 \times 1 0^{- 4} m o l L^{- 1}

Explanation:

$α = \frac{{[Λ]}_{M}^{C}}{{[Λ]}_{M}^{\infty}} = \frac{{[Λ]}_{M}^{C}}{{[Λ]}_{(H^{+})}^{\infty} {[Λ]}_{C H_{3} C O O^{-}}^{\infty}} = \frac{20}{350 + 50}$
$= \frac{20}{400} = \frac{1}{20}$
$K_{a} = \frac{C α^{2}}{1 - α}$
$K_{a} = C α^{2}$
$= 0 .007 \times {(\frac{1}{20})}^{2} = \frac{7 \times 1 0^{- 3}}{400}$
$K_{a} = \frac{7}{4} \times 1 0^{- 5}$
$= 1 .75 \times 1 0^{- 5} m o l L^{- 1}$

NEET - 2021

CHXII03:ELECTROCHEMISTRY

330328 If molar conductivity of ${Ca}^{2 +}$ and ${Cl}^{-}$ ions are 119 and $71 S {cm}^{2} {mol}^{- 1}$ respectively, then the molar conductivity of ${CaCl}_{2}$ at infinite dilution is

1

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

2

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

3

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

4

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

Explanation:

Given, $Λ_{C a^{2 +}}^{\circ} = 119 S c m^{2} mo l^{- 1}$
$Λ_{{Cl}^{-}}^{\circ} = 71 S {cm}^{2} {mol}^{- 1}$
As we know that,
$Λ_{{CaCl}_{2}}^{\circ} = Λ_{{Ca}^{2 +}}^{\circ} + 2 \times Λ_{{Cl}^{-}}^{\circ}$
$∴ Λ_{C a C l_{2}}^{\circ} = 119 + 2 \times 71$
$Λ_{C a C l_{2}}^{\circ} = 261 S c m^{2} m o l^{- 1}$

CHXII03:ELECTROCHEMISTRY

330329 According to Kohlrausch law, the limiting value of molar conductance of an electrolyte $A_{2} B$ is

1

λ_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}

2

{2 λ}_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}

3

λ_{A^{+}}^{^{\circ}} - λ_{B^{- 2}}^{^{\circ}}

4

λ_{A^{+}}^{^{\circ}} + \frac{1}{2} λ_{B^{- 2}}^{^{\circ}}

Explanation:

$Λ_{A_{2} B}^{^{\circ}} {= 2 λ}_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}$

CHXII03:ELECTROCHEMISTRY

330330 At Infinite dilution, each ion makes definite contribution to conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte. This law states that

1 Kohlrausch’s law

2 Debye Huckel rule

3 Faraday’s law

4 Arrhenius law

CHXII03:ELECTROCHEMISTRY

330331 The resistance of $0.1 M$ weak acid HA in a conductivity cell is $2 \times 10^{3} Ohm$ . The cell constant of the cell is $0.78 {cm}^{- 1}$ and $Λ_{m}^{\circ}$ of acid HA is $390 {Scm}^{2} {mol}^{- 1}$ . The $pH$ of the solution is

1 5

2 3

3 3.3

4 4.2

Explanation:

$C = 0.1 M; Λ_{m}^{0} = 390 {Scm}^{2} {mol}^{- 1}$
$\begin{aligned} R = 2 \times 10^{3} ohm \\ G^{*} = 0.78 {cm}^{- 1} \\ K = \frac{R}{G^{*}} = \frac{0.78}{2 \times 10^{3}} = 3.9 \times 10^{- 4} \\ Λ_{m} = \frac{K \times 1000}{C} = \frac{3.9 \times 10^{- 4} \times 1000}{0.1} = 3.9 \\ α = \frac{Λ_{m}}{Λ_{C}^{0}} = \frac{3.9}{390} = 10^{- 2} \\ [H^{+}] = c α = 0.1 \times 10^{- 2} = 10^{- 3} \\ pH = - \log 10^{- 3} = 3 \end{aligned}$

KCET - 2023

CHXII03:ELECTROCHEMISTRY

330333 The molar conductivity of 0.007 M acetic acid is $20 S . c m^{2} m o l^{- 1}$ .
What is the dissociation constant of acetic acid ? Choose the correct option.
$[\begin{array}{l} Λ_{H^{+}}^{o} = 350 S c m^{2} m o l^{- 1}; \\ Λ_{C H_{3} C O O^{-}}^{o} = 50 S c m^{2} m o l^{- 1} \end{array}]$

1

2 .50 \times 1 0^{- 4} m o l L^{- 1}

2

1 .75 \times 1 0^{- 5} m o l L^{- 1}

3

2 .50 \times 1 0^{- 5} m o l L^{- 1}

4

1 .75 \times 1 0^{- 4} m o l L^{- 1}

Explanation:

$α = \frac{{[Λ]}_{M}^{C}}{{[Λ]}_{M}^{\infty}} = \frac{{[Λ]}_{M}^{C}}{{[Λ]}_{(H^{+})}^{\infty} {[Λ]}_{C H_{3} C O O^{-}}^{\infty}} = \frac{20}{350 + 50}$
$= \frac{20}{400} = \frac{1}{20}$
$K_{a} = \frac{C α^{2}}{1 - α}$
$K_{a} = C α^{2}$
$= 0 .007 \times {(\frac{1}{20})}^{2} = \frac{7 \times 1 0^{- 3}}{400}$
$K_{a} = \frac{7}{4} \times 1 0^{- 5}$
$= 1 .75 \times 1 0^{- 5} m o l L^{- 1}$

NEET - 2021

CHXII03:ELECTROCHEMISTRY

330328 If molar conductivity of ${Ca}^{2 +}$ and ${Cl}^{-}$ ions are 119 and $71 S {cm}^{2} {mol}^{- 1}$ respectively, then the molar conductivity of ${CaCl}_{2}$ at infinite dilution is

1

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

2

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

3

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

4

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

Explanation:

Given, $Λ_{C a^{2 +}}^{\circ} = 119 S c m^{2} mo l^{- 1}$
$Λ_{{Cl}^{-}}^{\circ} = 71 S {cm}^{2} {mol}^{- 1}$
As we know that,
$Λ_{{CaCl}_{2}}^{\circ} = Λ_{{Ca}^{2 +}}^{\circ} + 2 \times Λ_{{Cl}^{-}}^{\circ}$
$∴ Λ_{C a C l_{2}}^{\circ} = 119 + 2 \times 71$
$Λ_{C a C l_{2}}^{\circ} = 261 S c m^{2} m o l^{- 1}$

CHXII03:ELECTROCHEMISTRY

330329 According to Kohlrausch law, the limiting value of molar conductance of an electrolyte $A_{2} B$ is

1

λ_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}

2

{2 λ}_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}

3

λ_{A^{+}}^{^{\circ}} - λ_{B^{- 2}}^{^{\circ}}

4

λ_{A^{+}}^{^{\circ}} + \frac{1}{2} λ_{B^{- 2}}^{^{\circ}}

Explanation:

$Λ_{A_{2} B}^{^{\circ}} {= 2 λ}_{A^{+}}^{^{\circ}} {+ λ}_{B^{- 2}}^{^{\circ}}$

CHXII03:ELECTROCHEMISTRY

330330 At Infinite dilution, each ion makes definite contribution to conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte. This law states that

1 Kohlrausch’s law

2 Debye Huckel rule

3 Faraday’s law

4 Arrhenius law

CHXII03:ELECTROCHEMISTRY

330331 The resistance of $0.1 M$ weak acid HA in a conductivity cell is $2 \times 10^{3} Ohm$ . The cell constant of the cell is $0.78 {cm}^{- 1}$ and $Λ_{m}^{\circ}$ of acid HA is $390 {Scm}^{2} {mol}^{- 1}$ . The $pH$ of the solution is

1 5

2 3

3 3.3

4 4.2

Explanation:

$C = 0.1 M; Λ_{m}^{0} = 390 {Scm}^{2} {mol}^{- 1}$
$\begin{aligned} R = 2 \times 10^{3} ohm \\ G^{*} = 0.78 {cm}^{- 1} \\ K = \frac{R}{G^{*}} = \frac{0.78}{2 \times 10^{3}} = 3.9 \times 10^{- 4} \\ Λ_{m} = \frac{K \times 1000}{C} = \frac{3.9 \times 10^{- 4} \times 1000}{0.1} = 3.9 \\ α = \frac{Λ_{m}}{Λ_{C}^{0}} = \frac{3.9}{390} = 10^{- 2} \\ [H^{+}] = c α = 0.1 \times 10^{- 2} = 10^{- 3} \\ pH = - \log 10^{- 3} = 3 \end{aligned}$

KCET - 2023

CHXII03:ELECTROCHEMISTRY

330333 The molar conductivity of 0.007 M acetic acid is $20 S . c m^{2} m o l^{- 1}$ .
What is the dissociation constant of acetic acid ? Choose the correct option.
$[\begin{array}{l} Λ_{H^{+}}^{o} = 350 S c m^{2} m o l^{- 1}; \\ Λ_{C H_{3} C O O^{-}}^{o} = 50 S c m^{2} m o l^{- 1} \end{array}]$

1

2 .50 \times 1 0^{- 4} m o l L^{- 1}

2

1 .75 \times 1 0^{- 5} m o l L^{- 1}

3

2 .50 \times 1 0^{- 5} m o l L^{- 1}

4

1 .75 \times 1 0^{- 4} m o l L^{- 1}

Explanation:

$α = \frac{{[Λ]}_{M}^{C}}{{[Λ]}_{M}^{\infty}} = \frac{{[Λ]}_{M}^{C}}{{[Λ]}_{(H^{+})}^{\infty} {[Λ]}_{C H_{3} C O O^{-}}^{\infty}} = \frac{20}{350 + 50}$
$= \frac{20}{400} = \frac{1}{20}$
$K_{a} = \frac{C α^{2}}{1 - α}$
$K_{a} = C α^{2}$
$= 0 .007 \times {(\frac{1}{20})}^{2} = \frac{7 \times 1 0^{- 3}}{400}$
$K_{a} = \frac{7}{4} \times 1 0^{- 5}$
$= 1 .75 \times 1 0^{- 5} m o l L^{- 1}$

NEET - 2021

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