Question

The molar conductivity of $0.025 \,mol\, L^{-1}$ methanoic acid is $46.1\, S \,cm^2 \,mol^{-1}$. If $\lambda(H^+) = 349.6 \,S \,cm^2 \,mol^{-1}$ and $\lambda(HCOO^-) = 54.6 \,S \,cm^2 \,mol^{-1}$ its degree of dissociation is

• A. 34%
• B. 25%
• C. 5%
• D. 11.4%

Answer 1

Answer: (D)

$\Lambda^{\circ}_{\left(HCOOH\right)} = \lambda^{\circ}_{\left(H^{+}\right)} + \lambda^{\circ}_{\left(HCOO^{-}\right)} $
$ =349.6 + 54.6 = 404 .2\, S \,cm^{2}\, mol^{-1} $
$\alpha = \frac{\Lambda_{m}}{\Lambda^{\circ}_{m}} = \frac{46.1}{404.2} = 0.114$
$\Rightarrow \alpha = 11.4 \%$