(B) The degree of dissociation $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$.Given $K_a = 1.8 \times 10^{-5}$ and $C = 0.01 \ M = 1

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(B) The degree of dissociation $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$.Given $K_a = 1.8 \times 10^{-5}$ and $C = 0.01 \ M = 10^{-2} \ M$.$\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{10^{-2}}} = \sqrt{1.8 \times 10^{-3}} = \sqrt{18 \times 10^{-4}} = 4.24 \times 10^{-2} = 0.0424$.The molar conductivity $\Lambda_m$ is related to the molar conductivity at infinite dilution $\Lambda_m^\circ$ by $\Lambda_m = \alpha \times \Lambda_m^\circ$.$\Lambda_m = 0.0424 \times 390 \ S \ cm^2 \ mol^{-1} = 16.536 \ S \ cm^2 \ mol^{-1} \approx 16.54 \ S \ cm^2 \ mol^{-1}$.

Class 12 Chemistry Electrochemistry Mix Examples-Electrochemistry

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The molar conductivity of acetic acid solution at infinite dilution is $390 \ S \ cm^2 \ mol^{-1}$. What is the molar conductivity of $0.01 \ M$ acetic acid solution (in $S \ cm^2 \ mol^{-1}$)? (Given: $K_{a}(CH_3COOH) = 1.8 \times 10^{-5}$,assume $1-\alpha \approx 1$)

TS EAMCET 2025 All TS EAMCET

A
$10.64$
B
$16.54$
C
$51.64$
D
$15.64$

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$1000 \, mL$ of $1 \, M$ $CuSO_{4(aq)}$ is electrolysed by $9.65 \, A$ current for $100 \, s$ using $Pt$ electrodes. Which statement is incorrect?

Medium

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The values of the limiting molar conductivity $(\lambda^0)$ for $NaCl$,$HCl$ and $NaOAc$ are $126.4$,$425.9$ and $91.0 \, S \, cm^2 \, mol^{-1}$,respectively. For $HOAc$,$\Lambda^0$ in $S \, cm^2 \, mol^{-1}$ is

MediumKVPY 2011

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Match List-$I$ with List-$II$.

List-$I$ List-$II$

$A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$ $I$. Primary battery

$B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$ $II$. Discharging of secondary battery

$C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$ $III$. Fuel cell

$D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$ $IV$. Charging of secondary battery

Choose the correct answer from the options given below.

MediumJEE MAIN 2022

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At $300 \ K$,the conductivity of $0.01 \ mol \ dm^{-3}$ aqueous solution of acetic acid is $19.5 \times 10^{-5} \ S \ cm^{-1}$ and the limiting molar conductivity of acetic acid at the same temperature is $390 \ S \ cm^2 \ mol^{-1}$. The degree of dissociation of acetic acid is:

MediumTS EAMCET 2023

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Given below are the half-cell reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn; E^{o} = -1.18 \ V$
$2(Mn^{3+} + e^{-} \rightarrow Mn^{2+}); E^{o} = +1.51 \ V$
The $E^{o}$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ will be:

DifficultJEE MAIN 2014

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List-$I$	List-$II$
$A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$	$I$. Primary battery
$B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$	$II$. Discharging of secondary battery
$C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$	$III$. Fuel cell
$D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$	$IV$. Charging of secondary battery

The molar conductivity of acetic acid solution at infinite dilution is $390 \ S \ cm^2 \ mol^{-1}$. What is the molar conductivity of $0.01 \ M$ acetic acid solution (in $S \ cm^2 \ mol^{-1}$)? (Given: $K_{a}(CH_3COOH) = 1.8 \times 10^{-5}$,assume $1-\alpha \approx 1$)

Similar Questions

$1000 \, mL$ of $1 \, M$ $CuSO_{4(aq)}$ is electrolysed by $9.65 \, A$ current for $100 \, s$ using $Pt$ electrodes. Which statement is incorrect?

The values of the limiting molar conductivity $(\lambda^0)$ for $NaCl$,$HCl$ and $NaOAc$ are $126.4$,$425.9$ and $91.0 \, S \, cm^2 \, mol^{-1}$,respectively. For $HOAc$,$\Lambda^0$ in $S \, cm^2 \, mol^{-1}$ is

At $300 \ K$,the conductivity of $0.01 \ mol \ dm^{-3}$ aqueous solution of acetic acid is $19.5 \times 10^{-5} \ S \ cm^{-1}$ and the limiting molar conductivity of acetic acid at the same temperature is $390 \ S \ cm^2 \ mol^{-1}$. The degree of dissociation of acetic acid is:

Given below are the half-cell reactions:$Mn^{2+} + 2e^{-} \rightarrow Mn; E^{o} = -1.18 \ V$$2(Mn^{3+} + e^{-} \rightarrow Mn^{2+}); E^{o} = +1.51 \ V$The $E^{o}$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ will be:

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Given below are the half-cell reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn; E^{o} = -1.18 \ V$
$2(Mn^{3+} + e^{-} \rightarrow Mn^{2+}); E^{o} = +1.51 \ V$
The $E^{o}$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ will be: