Question

The specific conductance of $0.01\text{M}$ solution of acetic was found to be $0.0163{\text{S m}}^{-1}$ at $25{}^{\circ }\text{C}$. Calculate the degree of dissociation of the acid.
Molar conductance of acetic acid at infinite dilution is $652\times {10}^{-4}{\text{S m}}^2{\text{mol}}^{-1}$ at $25{}^{\circ }\text{C}$.

• A. $0.05$
• B. $0.025$
• C. $0.42$
• D. $0.069$

Answer 1

The correct option is B $0.025$

Given:
$\text{Conductivity,}\kappa =0.0163{\text{S m}}^{-1}$

Also, we know
$1\text{L}={10}^{-3}{\text{m}}^3$
$1{\text{L}}^{-1}={10}^3{\text{m}}^{-3}$
$\text{Molar concentration,}\left(C\right)=0.01{\text{mol L}}^{-1}=10{\text{mol m}}^{-3}$

$\text{Molar conductance,}{\Lambda }_m=\frac{\kappa }{C}$
${\Lambda }_m=\frac{0.0163{\text{S m}}^{-1}}{10{\text{mol m}}^{-3}}=16.3\times {10}^{-4}{\text{S m}}^2{\text{mol}}^{-1}$

Given,
${\Lambda }_m^0=652\times {10}^{-4}{\text{S m}}^2{\text{mol}}^{-1}$

$\text{Degree of dissociation,}\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}=\frac{16.3\times {10}^{-4}{\text{S m}}^2{\text{mol}}^{-1}}{652\times {10}^{-4}{\text{S m}}^2{\text{mol}}^{-1}}=0.025$