At 298 K,the vapour pressure of a solution o | vedclass

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(C) Key Idea: The relative lowering of vapour pressure is given by $\frac{p^{\circ} - p}{p^{\circ}} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} =

Vedclass · Accepted Answer

$71.5$

Vedclass · Answer

(C) Key Idea: The relative lowering of vapour pressure is given by $\frac{p^{\circ} - p}{p^{\circ}} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{w_2 / M_2}{w_1 / M_1}$.Given:Weight of non-volatile solute,$w_2 = 7.5 \ g$.In the first case,weight of water,$w_1 = 90 \ g$,vapour pressure of solution $p_1 = 2.8 \ kPa$.$\frac{p^{\circ} - 2.8}{p^{\circ}} = \frac{7.5 / M_2}{90 / 18} = \frac{7.5}{M_2 	imes 5} = \frac{1.5}{M_2} \quad \dots (i)$In the second case,weight of water,$w_1' = 90 + 18 = 108 \ g$,vapour pressure of solution $p_2 = 2.81 \ kPa$.$\frac{p^{\circ} - 2.81}{p^{\circ}} = \frac{7.5 / M_2}{108 / 18} = \frac{7.5}{M_2 	imes 6} = \frac{1.25}{M_2} \quad \dots (ii)$Subtracting $(ii)$ from $(i)$:$\frac{p^{\circ} - 2.8}{p^{\circ}} - \frac{p^{\circ} - 2.81}{p^{\circ}} = \frac{1.5}{M_2} - \frac{1.25}{M_2}$$\frac{0.01}{p^{\circ}} = \frac{0.25}{M_2} \implies p^{\circ} = \frac{0.01 M_2}{0.25} = 0.04 M_2$Substituting $p^{\circ}$ in $(i)$:$1 - \frac{2.8}{0.04 M_2} = \frac{1.5}{M_2}$$1 = \frac{1.5}{M_2} + \frac{70}{M_2} = \frac{71.5}{M_2}$$M_2 = 71.5 \ g \ mol^{-1}$.

At $298 \ K$,the vapour pressure of a solution of $7.5 \ g$ of non-volatile solute in $90 \ g$ of water is $2.8 \ kPa$. If $18 \ g$ of water is added to this solution,the vapour pressure becomes $2.81 \ kPa$ at the same temperature. The molar mass of the solute in $g \ mol^{-1}$ is:

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