Question

The equivalent conductance of $0.02M$ acetic acid is $1.162\times {10}^{-3}$ ohm${}^{-1}$ $mo{l}^{-1}$ $m^2$. Degree of ionisation '$\alpha$' of $C{H}_{3}COOH$ is : $({\lambda }_{H^{+}}=349.83oh{m}^{-1}and{\lambda }_{C{H}_{3}CO{O}^{-}}=40.89oh{m}^{-1})$

• A. $0.01$
• B. $0.02$
• C. $0.03$
• D. $0.04$

Answer 1

Answer: (C)

$0.03$

The equivalent conductance of an electrolyte at infinite dilution is equal to the sum of contribution of the individual ions. ${\Lambda }_{eq}^{\infty }=349.8+40.89=390.72$.
The degree of dissociation is the ratio of the equivalent conductance at dilution 'v' to the equivalent conductance at infinite dilution. $\alpha =\frac{{\Lambda }_{eq}^v}{{\Lambda }_{eq}^{\infty }}=\frac{1.162\times {10}^{-3}}{390.72}=0.003\times {10}^{-3}$.