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Q.
The dissociation constant of $n$-butyric acid is $1.6 \times 10^{-5}$ and the molar conductivity at infinite dilution is $380 \times 10^{-4} Sm ^{2}\, mol ^{-1}$. The specific conductance of the $0.01\, M$ acid solution is :-
Solution:
$K _{ a }= C \alpha^{2}$
$\alpha=\sqrt{ K _{ a } / C }$
$\alpha=\sqrt{\frac{1.6 \times 10^{-5}}{0.01}}$
$\alpha=0.04$
$\alpha=\frac{\lambda m }{\lambda_{ m } \infty}$
$\lambda_{ m }=0.04 \times 380 \times 10^{-4}$
$=15.2 \times 10^{-4} \,S m ^{2}\, mol ^{-1}$
or
$=15.2\, S\, cm ^{2}\, mol ^{-1}$
$\lambda_{ m }=\frac{1000 \times K }{ M }$
$K =\frac{15.2 \times 0.01}{1000}$
$K =152 \times 10^{-6}$
$K =1.52 \times 10^{-4} S\, cm ^{-1}$
or $1.52 \times 10^{-2} S\, m ^{-1}$