At 50^{circ} C, the vapour pressure of pure b | vedclass

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(C) According to Raoult's law for a solution containing a non-volatile solute, the vapour pressure of the solution is given by $P = P^{\circ} \cdot x_

Vedclass · Accepted Answer

$0.605$

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(C) According to Raoult's law for a solution containing a non-volatile solute, the vapour pressure of the solution is given by $P = P^{\circ} \cdot x_{	ext{solvent}}$.Here, $P = 167 \ torr$ and $P^{\circ} = 268 \ torr$.So, the mole fraction of benzene is $x_{	ext{benzene}} = \frac{P}{P^{\circ}} = \frac{167}{268} \approx 0.6231$.Since $x_{	ext{benzene}} = \frac{n_{	ext{benzene}}}{n_{	ext{benzene}} + n_{	ext{solute}}}$, we set $n_{	ext{benzene}} = 1 \ mol$.Then, $0.6231 = \frac{1}{1 + n_{	ext{solute}}}$.$1 + n_{	ext{solute}} = \frac{1}{0.6231} \approx 1.6048$.$n_{	ext{solute}} = 1.6048 - 1 = 0.6048 \ mol \approx 0.605 \ mol$.

At $50^{\circ} C$, the vapour pressure of pure benzene is $268 \ torr$. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of $167 \ torr$ at the same temperature is (molar mass of benzene $= 78 \ g \ mol^{-1}$)

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