Question

The specific conductance $(k)$ of $0.02M$ aqueous acetic acid solution a t $298K$ is $1.65\times 10^{-4}Sc{m}^{-1}$.The degree of dissociation of acetic acid is [Given equivalent conductance at infinite dilution of $H^{-}=349.1Sc{m}^{2}mo{l}^{-1}$ and $C{H}_{3}CO{O}^{-}=40.9Sc{m}_{2}mo{l}_{-1}$]

• A. 0.021
• B. 0.21
• C. 0.012
• D. 0.12

Answer 1

Answer: (A)

Given, specific conductance of acetic acid $=1.65\times 10^{-4}Sc{m}^{-2}$
Molarity of acetic acid $=0.02M$
As we know, molar conductance can be calculated using formula
${\lambda }_m=\frac{1000\times k}{M}$
${\lambda }_{m_{C{H}_{3}COOH}}=\frac{1000\times 1.65\times 10^{-4}}{0.02}=8.25$
${\lambda }_{m_{C{H}_{3}COOH}}^{\infty }={\lambda }_{m\left(H^{+}\right)}^{\infty }+{\lambda }_{m\left(C{H}_{3}CO{O}^{-}\right)}^{\infty }$
$=(349.1+40.9)Sc{m}^{2}mo{l}^{-1}$
$=390$
Degree of dissociation, $\alpha =\frac{{\lambda }_{mC{H}_{3}COOH}}{{\lambda }_{mC{H}_{3COOH}}^{\infty }}$
$\alpha =\frac{8.25}{390}=0.0211$