JEE Advance - Chemistry (2018 - Paper 1 Offline - No. 14)
On addition of equal number of moles of a non-volatile solute $$S$$ in equal amount (in $$kg$$) of these solvents, the elevation of boiling point of solvent $$X$$ is three times that of solvent $$Y$$. Solute $$S$$ is known to undergo dimerization in these solvents. If the degree of dimerization is $$0.7$$ in solvent $$Y$$, the degree of dimerization in solvent $$X$$ is ___________.
Explanation
For solvent X, $$\Delta$$TbX = KbX m $$\Rightarrow$$ 2 = 2KbX m ..... (1)
For solvent Y, $$\Delta$$TbY = KbY m $$\Rightarrow$$ 1 = 2KbY 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
$${{\Delta {T_{bX}}} \over {\Delta {T_{bY}}}} = {{\left( {1 - {\alpha \over 2}} \right){K_{bX}}m} \over {\left( {{{1.3} \over 2}} \right){K_{bY}}m}} \Rightarrow {1 \over 3} = {{2(2 - \alpha )} \over {1.3}} \Rightarrow \alpha = 0.05$$
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