Question

The specific conductance of $0.01M$ solution of acetic acid was found to be $0.0163S{m}^{-1}$ at $25{}^{o}C$. Calculate the degree of dissociation of the acid.
Molar conductance of acetic acid at infinite dilution is $390.7\times {10}^{-4}S{m}^{2}mo{l}^{-1}$ at $25{}^{o}C$.

• A. $0.0417$
• B. $0.0823$
• C. $0.0632$
• D. $0.0984$

Answer 1

The correct option is A $0.0417$

Given,
$\kappa =0.0163S{m}^{-1};c=0.01mold{m}^{-3}=10mol{m}^{-3}$

${\Lambda }_m=\frac{\kappa }{c}=\frac{0.0163S{m}^{-1}}{10mol{m}^{-3}}=16.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\circ }=390.7\times {10}^{-4}S{m}^{2}mo{l}^{-1}\text{(given)}$

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{16.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}}{390.7\times {10}^{-4}S{m}^{2}mo{l}^{-1}}=0.0417$