Solutions and Colligative Properties - Result Question 20
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.
$(1996,3 M)$
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Answer:
Correct Answer: 19. $(0.72)$
Solution:
Volume of $1.0$ mole liquid benzene $=\frac{78}{0.877} mL=88.94$ $ mL$
$\Rightarrow$ Molar volume of benzene vapour at $20^{\circ} C$
$=\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$
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 $=\chi$
$\Rightarrow 4.65 \chi+22.3(1-\chi) =45 $
$\Rightarrow \chi =0.43$
$\Rightarrow$ 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} $
BETA
