JEE Advance - Chemistry (2019 - Paper 1 Offline - No. 14)
On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapor pressure decreases from 650 mmHg to 640 mmHg. The depression of freezing point of benzene (in K) upon addition of the solute is .............
(Given data: Molar mass and the molal freezing point depression constant of benzene are 78 g mol-1 and 5.12 K kg mol-1, respectively).
(Given data: Molar mass and the molal freezing point depression constant of benzene are 78 g mol-1 and 5.12 K kg mol-1, respectively).
Answer
1.02
Explanation
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 (ps) = 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,
$$\Delta {T_f} = {K_f} \times molality$$
$$ = {{{K_f} \times {w_{solute}}} \over {{{(MW)}_{solute}} \times {w_{solvent}}}}$$
$$ \Rightarrow $$ $$ \Delta {T_f} = 5.12 \times {{0.5 \times 1000} \over {64 \times 39}} $$
$$\Rightarrow \Delta {T_f} = 1.02K$$
Given, vapour pressure of solvent (p$$^\circ $$) = 650 mmHg
Vapour pressure of solution (ps) = 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,
$$\Delta {T_f} = {K_f} \times molality$$
$$ = {{{K_f} \times {w_{solute}}} \over {{{(MW)}_{solute}} \times {w_{solvent}}}}$$
$$ \Rightarrow $$ $$ \Delta {T_f} = 5.12 \times {{0.5 \times 1000} \over {64 \times 39}} $$
$$\Rightarrow \Delta {T_f} = 1.02K$$
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