QJEEAdv19Ch8e28: JEE Advanced 2019 Chemistry - Class 12 JEEAdv19Ch8e28 Solution: On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g…

When 0.5 g of non-volatile solute dissolve into 39 gm of benzene then relative lowering of vapour pressure occurs. Hence, vapour pressure decreases from 650 mmHg to 640 mmHg.

Given, vapour pressure of solvent (p $^\circ$ ) = 650 mmHg

Vapour pressure of solution (p_s) = 640 mmHg

Weight of non-volatile solute = 0.5 g

Weight of solvent (benzene) = 39 g

From relative lowering of vapour pressure,

${{p^\circ - {p_s}} \over {p^\circ }} = {x_{Solute}} = {{{n_{solute}}} \over {{n_{solute}} + {n_{solvent}}}}$

${{650 - 640} \over {650}} = {{{{0.5} \over {molar\,mass}}} \over {{{0.5} \over {molar\,mass}} + {{39} \over {78}}}}$

${{10} \over {650}} = {{{{0.5} \over {molar\,mass}}} \over {{{0.5} \over {molar\,mass}} + 0.5}}$

$0.5 + 0.5 \times molar\,mass = 65 \times 0.5$

$\therefore$ Molar mass of solute = 64 g

From molal depression of freezing point,

$\Delta {T_f} = {K_f} \times molality$

$= {{{K_f} \times {w_{solute}}} \over {{{(MW)}_{solute}} \times {w_{solvent}}}}$

$\Rightarrow$ $\Delta {T_f} = 5.12 \times {{0.5 \times 1000} \over {64 \times 39}}$

$\Rightarrow \Delta {T_f} = 1.02K$