$38 \cdot 4 \mathrm{~g}$ of unknown substance (molar mass $384 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and $116…
$38 \cdot 4 \mathrm{~g}$ of unknown substance (molar mass $384 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and $116 \mathrm{~g}$ of acetone is used to prepare a solution at $313 \mathrm{~K}$. If vapour pressure of pure acetone (molar mass $58 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is $0.842$ atmosphere, what is the vapour pressure of solution?
$0.650 \mathrm{~atm}$
$0.880 \mathrm{~atm} .$
$0.7999 \mathrm{~atm} .$
$0.958 \mathrm{~atm}$
Solution
$\mathrm{n}_{2}=\frac{38.4}{384}=0.1 \mathrm{~mol}$ solute
$\mathrm{n}_{1}=\frac{116}{58}=2 \mathrm{~mol}$ solvent (acetone)
Mole fraction $; x_{1}=\frac{n_{1}}{n_{1}+n_{2}}=\frac{2}{2+0.1}=0.95$
By Raoult's law, the vapour pressure of the solution is given by
$P=x_{1} \times P_{0}=0.95 \times 0.842 \mathrm{~atm}=0.7999 \mathrm{~atm}$