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

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(A) $1$. Calculate moles of components:$n_{	ext{benzene}} = \frac{23.4 \ g}{78 \ g/mol} = 0.3 \ mol$$n_{	ext{toluene}} = \frac{64.4 \ g}{92 \ g/mol}

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$0.406$

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(A) $1$. Calculate moles of components:$n_{	ext{benzene}} = \frac{23.4 \ g}{78 \ g/mol} = 0.3 \ mol$$n_{	ext{toluene}} = \frac{64.4 \ g}{92 \ g/mol} = 0.7 \ mol$$2$. Calculate mole fractions in liquid phase $(x)$:$x_{	ext{benzene}} = \frac{0.3}{0.3 + 0.7} = 0.3$$x_{	ext{toluene}} = \frac{0.7}{0.3 + 0.7} = 0.7$$3$. Calculate partial pressures using Raoult's Law:$P_{	ext{benzene}} = x_{	ext{benzene}} 	imes P^0_{	ext{benzene}} = 0.3 	imes 75 = 22.5 \ mm \ Hg$$P_{	ext{toluene}} = x_{	ext{toluene}} 	imes P^0_{	ext{toluene}} = 0.7 	imes 22 = 15.4 \ mm \ Hg$$4$. Calculate total pressure $(P_{	ext{total}})$:$P_{	ext{total}} = 22.5 + 15.4 = 37.9 \ mm \ Hg$$5$. Calculate mole fraction of toluene in vapour phase $(y_{	ext{toluene}})$:$y_{	ext{toluene}} = \frac{P_{	ext{toluene}}}{P_{	ext{total}}} = \frac{15.4}{37.9} \approx 0.406$

List-$I$	List-$II$
$A$. van't Hoff factor,$i$	$I$. Cryoscopic constant
$B$. $k_{f}$	$II$. Isotonic solutions
$C$. Solutions with same osmotic pressure	$III$. $\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$
$D$. Azeotropes	$IV$. Solutions with same composition of vapour above it

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