Solutions and Colligative Properties 1 Question 20
19. The molar volume of liquid benzene (density $=0.877 \mathrm{~g} / \mathrm{mL}$ ) increases by a factor of 2750 as it vaporises at $20^{\circ} \mathrm{C}$ and that of liquid toluene (density $=0.867 \mathrm{~g} \mathrm{~mL}^{-1}$ ) increases by a factor of 7720 at $20^{\circ} \mathrm{C}$. A solution of benzene and toluene at $20^{\circ} \mathrm{C}$ has a vapour pressure of 45.0 torr. Find the mole fraction of benzene in the vapour above the solution.
$(1996,3 \mathrm{M})$
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Solution:
- Volume of 1.0 mole liquid benzene $=\frac{78}{0.877} \mathrm{~mL}=88.94 \mathrm{~mL}$
$\Rightarrow$ Molar volume of benzene vapour at $20^{\circ} \mathrm{C}$
$$ \begin{aligned} =\frac{88.94 \times 2750}{1000} \mathrm{~L} & =244.58 \mathrm{~L} \ \Rightarrow \mathrm{VP} \text { of pure benzene at } 20^{\circ} \mathrm{C} & =\frac{0.082 \times 293}{244.58} \times 760 \mathrm{~mm} \ & =74.65 \mathrm{~mm} \end{aligned} $$
Similarly; molar volume of toluene vapour
$$ =\frac{92}{0.867} \times \frac{7720}{1000} \mathrm{~L}=819.2 \mathrm{~L} $$
$\Rightarrow$ VP of pure toluene $=\frac{0.082 \times 293}{819.2} \times 760 \mathrm{~mm}=22.3 \mathrm{~mm}$
Now, let mole fraction of benzene in the liquid phase $=\chi$
$$ \begin{aligned} \Rightarrow & & 4.65 \chi+22.3(1-\chi) & =45 \ \Rightarrow & & \chi & =0.43 \end{aligned} $$
$\Rightarrow$ Mole fraction of benzene in vapour phase
$$ \begin{aligned} & =\frac{\text { Partial vapour pressure of benzene }}{\text { Total vapour pressure }} \ & =\frac{74.65 \times 0.43}{45}=0.72 \end{aligned} $$
