Question

$k=4.95 \times 10^{-5} \,S \,cm ^{-1}$ for a $0.001\, M$ solution. The reciprocal of the degree of dissociation of acetic acid, if $\Lambda_{ m }{ }^{\circ}$ for acetic acid is $400\, S\, cm ^{-2} mol ^{-1}$ is:

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

Answer: (B)

$\kappa=4.95 \times 10^{-5} S cm ^{-1} ; \alpha=\frac{\Lambda_{ m }}{\Lambda_{ m }^0}$
$=1000 \times \frac{\kappa}{C}=1000 \times \frac{4.95 \times 10^{-5}}{0.001}=49.5\, S\, cm ^{-1}$
$\Rightarrow \alpha=\frac{49.5}{400} \Rightarrow \frac{1}{\alpha}=\frac{400}{49.5} \approx 8$