Question

At ${25}^{\circ }\text{C}$, the specific conductance of $0.01\text{M}$ aqueous solution of acetic acid is $1.63\times {10}^{-2}{\text{S m}}^{-1}$ and the molar conductance at infinite dilution is $390\times {10}^{-4}{\text{S m}}^2{\text{mol}}^{-1}$. Calculate the degree of dissociation.

• A. $0.041$
• B. $0.051$
• C. $0.023$
• D. $0.021$

Answer 1

The correct option is A $0.041$

$\text{Specific conductance, k}=1.63\times {10}^{-2}{\text{S m}}^{-1};c=0.01{\text{mol dm}}^{-3}=0.01\times {10}^3{\text{mol m}}^{-3}$
${\Lambda }_m=\frac{k}{c}=\frac{1.63\times {10}^{-2}{\text{S m}}^{-1}}{0.01\times {10}^3{\text{mol m}}^{-1}}=16.3\times {10}^{-4}{\text{S m}}^2{\text{mol}}^{-1}$
The degree of dissociation is given by:
$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{16.3\times {10}^{-4}{\text{S m}}^2{\text{mol}}^{-1}}{390\times {10}^{-4}{\text{S m}}^2{\text{mol}}^{-1}}=0.041$