Solutions and Colligative Properties - Result Question 48
####19. The molar volume of liquid benzene (density $=0.877 g / mL$ ) increases by a factor of 2750 as it vaporises at $20^{\circ} C$ and that of liquid toluene (density $=0.867 g mL^{-1}$ ) increases by a factor of 7720 at $20^{\circ} C$. A solution of benzene and toluene at $20^{\circ} C$ has a vapour pressure of 45.0 torr. Find the mole fraction of benzene in the vapour above the solution.
$(1996,3 M)$
Show Answer
Solution:
- Volume of 1.0 mole liquid benzene $=\frac{78}{0.877} mL=88.94 mL$
$\Rightarrow$ Molar volume of benzene vapour at $20^{\circ} C$
$$ \begin{aligned} =\frac{88.94 \times 2750}{1000} L & =244.58 L \ \Rightarrow VP \text { of pure benzene at } 20^{\circ} C & =\frac{0.082 \times 293}{244.58} \times 760 mm \ & =74.65 mm \end{aligned} $$
Similarly; molar volume of toluene vapour
$$ =\frac{92}{0.867} \times \frac{7720}{1000} L=819.2 L $$
$\Rightarrow$ VP of pure toluene $=\frac{0.082 \times 293}{819.2} \times 760 mm=22.3 mm$
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
$$ \begin{aligned} & =\frac{\text { Partial vapour pressure of benzene }}{\text { Total vapour pressure }} \ & =\frac{74.65 \times 0.43}{45}=0.72 \end{aligned} $$
