JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
A solution containing \(10 \mathrm{~g}\) of an electrolyte \(\mathrm{AB}_2\) in \(100 \mathrm{~g}\) of water boils at \(100.52^{\circ} \mathrm{C}\). The degree of ionization of the electrolyte \((\alpha)\) is _______ \(\times 10^{-1}\). (nearest integer) [Given : Molar mass of \(\mathrm{AB}_2=200 \mathrm{~g} \mathrm{~mol}^{-1} \cdot \mathrm{K}_{\mathrm{b}}\) (molal boiling point elevation const. of water) \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\), boiling point of water \(=100^{\circ} \mathrm{C}\); \(\mathrm{AB}_2\) ionises as \(\left.\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}\right]\)
- A \(3\)
- B \(5\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
\(\mathrm{AB}_2 \rightarrow \mathrm{A}^{+2}+2 \mathrm{~B}^{\ominus}\) \(\mathrm{i}=1+(3-1) \alpha\) \(\mathrm{i}=1+2 \alpha\) \(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{k}_{\mathrm{b}} \mathrm{im}\) \(0.52=0.52(1+2 \alpha) \frac{\frac{10}{200}}{\frac{100}{1000}}\)…
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