Chemistry - Liquid Solution Question with Solution | TestHub

$\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{S}{\mathrm{O}}_4\to \underset{0.08}{\mathrm{N}{\mathrm{a}}^{+}}+\underset{0.08}{\mathrm{H}\mathrm{S}{\mathrm{O}}_4^{-}}$

$\underset{0.08(1-\mathrm{\alpha })}{\mathrm{H}\mathrm{S}{\mathrm{O}}_4^{-}}\rightleftharpoons \underset{0.08\mathrm{\alpha }}{{\mathrm{H}}^{+}}+\underset{0.08\mathrm{\alpha }}{\mathrm{S}{\mathrm{O}}_4^{2-}}$

$\mathrm{i}=\frac{0.08+0.08\left(1-\mathrm{\alpha }\right)+0.08\mathrm{\alpha }+0.08\mathrm{\alpha }}{0.08}=2+\mathrm{\alpha }$

$\mathrm{\Delta }\mathrm{T}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}$

$0.372=\mathrm{i}\times 1.86\times 0.08$

$\mathrm{i}=2.5$

So$2+\mathrm{\alpha }=2.5$

$\mathrm{\alpha }=0.5$

Dissociation constant,$\mathrm{K}=\frac{0.08\mathrm{\alpha }\times 0.08\mathrm{\alpha }}{0.08\left(1-\mathrm{\alpha }\right)}$

$=\frac{0.08\times 0.5\times 0.08\times 0.5}{0.08\times 0.5}=0.04$