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Liquid Solution

Section: Chemistry

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Question : 5 of 29

Marks: +1, -0

At

T(\mathrm{K})

, the vapour pressure of pure benzene and toluene are 75 and 22 mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed to form an ideal solution. If the vapours are in equilibrium with the liquid mixture, the mole fraction of toluene in vapour phase is (At.wt of

C=12 ; H=1

[AP EAPCET 23 May 2025 S1]

0.406
0.594
0.539
0.461

Solution:

Moles of benzene

=n_B=\frac{23.4}{78}=0.3\,\text{mol}

Moles of toluene

=n_{\text{rolu}}=\frac{64.4}{92}=0.7\,\text{mol}

Moles fraction,

\chi_{\text{Tolu}}=\frac{n_{\text{Tolu}}}{n_{\text{Total}}}=\frac{0.7}{1}=0.7

;

\chi_B=1-0.7=0.3

Partial pressure of toluene

p_{\text{Tolu}}=\chi_{\text{Total}} \cdot p_{\text{Tolu}}^{\circ}

\Rightarrow 0.7 \times 22=15.4\,\text{mmHg}

Partial pressure of benzene

p_{\text{Ben}}=\chi_B \cdot p_B

\Rightarrow 0.3 \times 75=225\,\text{mmHg}

Total vapour, pressure

=22.5+15.4=37.9\,\text{mmHg}

Mole fraction of toluene in vapour

Y_{\text{toluene}}=\frac{p_{\text{Toluene}}}{p_{\text{Total}}}=\frac{15.4}{37.9}=0.406

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