$100 \mathrm{~g}$ of liquid $\mathrm{A}\left(\right.$ molar mass $140 \mathrm{~g} \mathrm{~mol}^{-1}$ ) was dissolved in $1000 \mathrm{~g}$ of liquid $\mathrm{B}$ (molar mass $180 \mathrm{~g} \mathrm{~mol}^{-1}$ ). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid $A$ and its vapour pressure in the solution if the total vapour pressure of the solution is $\mathbf{4 7 5}$ torr.

Question

$100 \mathrm{~g}$ of liquid $\mathrm{A}\left(\right.$ molar mass $140 \mathrm{~g} \mathrm{~mol}^{-1}$ ) was dissolved in $1000 \mathrm{~g}$ of liquid $\mathrm{B}$ (molar mass $180 \mathrm{~g} \mathrm{~mol}^{-1}$ ). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid $A$ and its vapour pressure in the solution if the total vapour pressure of the solution is $\mathbf{4 7 5}$ torr.,

MARKS App · Accepted Answer

No. of moles of solute, $n_2=\frac{100}{140}=\frac{5}{7}$ mole No. of moles of solvent, $n_1=\frac{1000}{180}=\frac{50}{9}$ mole Mole fraction of solute, $x_2=\frac{n_2}{n_1+n_2}=\frac{5 / 7}{5 / 7+50 / 9}=0 \cdot 114$ Mole fraction of solvent, $x_1=\left(1-x_2\right)=(1-0 \cdot 114)=0.886$ According to Raoult's law $\begin{aligned} &P_A=x_A P_A^{\circ}=0.114 \times P_A^{\circ} \ &P_B=x_B P_B^{\circ}=0.886 \times 500=443 \text { torr } \ &P_{\text {Total }}=P_A+P_B \ &475=0.114 P_A^{\circ}+443 \end{aligned}$ $\begin{aligned} &P_A^{\circ}=\frac{475-443}{0.114}=280 \cdot 7 \text { torr } \ &\therefore P_A=0.114 \times 280.7=32 \text { torr. } \end{aligned}$

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