QJEEAdv18Ch790f: JEE Advanced 2018 Chemistry - Class 12 JEEAdv18Ch790f Solution: The plot given below shows curves (where is the pressure and is …

For solvent X, $\Delta$ T_bX = K_bX m $\Rightarrow$ 2 = 2K_bX m ..... (1)

For solvent Y, $\Delta$ T_bY = K_bY m $\Rightarrow$ 1 = 2K_bY m ..... (2)

Dividing Eq. (1) by Eq. (2), we get

${{{K_{bX}}} \over {{K_{bY}}}} = 2$

After adding solute S, molality is the same for both solutions.

Dimerisation is a association property. And for dimerisation van't Hoff factor

i = 1 + $\left( {{1 \over n} - 1} \right)\alpha$

And for dimerisation n = 2

$\therefore$ For solvent X, van't Hoff factor $i = 1 - {\alpha \over 2}$

$\Delta {T_{bX}} = i{K_{bX}}m = \left( {1 - {\alpha \over 2}} \right){K_{bX}}m$ ....... (3)

For solvent Y, van't Hoff factor $i = 1 - {{0.7} \over 2} = {{1.3} \over 2}$

$\Delta {T_{bY}} = i{K_{bY}}m \Rightarrow \left( {{{1.3} \over 2}} \right){K_{bY}}m$ ....... (4)

Given that $\Delta {T_{bX}} = 3\Delta {T_{bY}}$ . Dividing Eq. (3)/(4), we have

${{\Delta {T_{bX}}} \over {\Delta {T_{bY}}}} = {{\left( {1 - {\alpha \over 2}} \right){K_{bX}}m} \over {\left( {{{1.3} \over 2}} \right){K_{bY}}m}} \Rightarrow {1 \over 3} = {{2(2 - \alpha )} \over {1.3}} \Rightarrow \alpha = 0.05$