Question

A solution containing 10g of an electrolyte \(AB_2\) in 100g of water boils at \(100.52^\circ C\). The degree of ionization of the electrolyte (\(\alpha\)) is _____ \(\times 10 ^{-1}.\) (nearest integer)

[Given : Molar mass of \(AB_2\) = 200g \(mol^{-1 }\). \(K_b\) (molal boiling point elevation const. of water) = 0.52 K kg \(mol^{-1}\) , boiling point of water = \(100 ^\circ C\) ; \(AB_2\) ionises as \(AB_2 \rightarrow A^{2+} +2B^{-}\)]

Answer 1

Correct answer is : 5

\(\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}\)

\(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{ik}_{\mathrm{b}} \mathrm{m}\)

\((0.52)=\mathrm{i} \times 0.52 \times \frac{\frac{10}{200}}{100 \times 10^{-3}}\)

\(\mathrm{i}=2\)

\(\mathrm{AB}_2 \rightarrow \mathrm{A}+2 \mathrm{~B} \)

\( \mathrm{i}=1+(3-1) \alpha\)

\(2=1+(3-1) \alpha\)

\( \frac{2-1}{3-1}=\alpha\)

\(\alpha=\frac{1}{2} \quad \alpha=0.5 \quad \Rightarrow 5 \times 10^{-1}\)