At T (K),the vapour pressure of pure benzene | vedclass

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(D) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by: $\frac{p^{\circ} - p}{p^{\circ}} = \frac{

Vedclass · Accepted Answer

$169$

Vedclass · Answer

(D) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by: $\frac{p^{\circ} - p}{p^{\circ}} = \frac{n_1}{n_1 + n_2} \approx \frac{n_1}{n_2} = \frac{w_1 	imes M_2}{M_1 	imes w_2}$.Given: $p^{\circ} = 0.85$ bar,$p = 0.845$ bar,$w_1 = 0.5$ $g$,$w_2 = 39$ $g$,$M_2$ (molar mass of benzene,$C_6H_6$) $= 78$ $g$ $mol^{-1}$.Substituting the values: $\frac{0.85 - 0.845}{0.85} = \frac{0.5 	imes 78}{M_1 	imes 39}$.$\frac{0.005}{0.85} = \frac{0.5 	imes 2}{M_1} = \frac{1}{M_1}$.$M_1 = \frac{0.85}{0.005} = 170$ $g$ $mol^{-1}$.Note: Using the formula $\frac{p^{\circ} - p}{p} = \frac{n_1}{n_2}$ as per the provided solution logic yields $M_1 = \frac{0.5 	imes 78 	imes 0.845}{0.005 	imes 39} = 169$ $g$ $mol^{-1}$. Thus,the correct option is $D$.

At $T$ $(K)$,the vapour pressure of pure benzene is $0.85$ bar. $A$ non-volatile,non-electrolyte substance weighing $0.5$ $g$ when added to $39$ $g$ of benzene,the vapour pressure of the solution is $0.845$ bar. The molar mass (in $g$ $mol^{-1}$) of the substance is

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