Q29 / 89 · KVPY 2017 (5 Nov SB/SX) · Chemistry

A mixture of toluene and benzene boils at $100^{\circ} \mathrm{C}$. Assuming ideal behaviour, the mole fraction of toluene in the mixture is closest to [Vapour pressures of pure toluene and pure benzene at $100^{\circ} \mathrm{C}$ are $0.742$ and $1.800$ bar respectively. 1 atm $=1.013$ bar $]$

A
$0.824$
Correct answer
B
$0.744$
C
$0.544$
D
$0.624$

Solution

$1.013=0.742 \mathrm{X}_{\mathrm{t}}+1.8 \mathrm{X}_{\mathrm{b}}$ $1.013=0.742\left(1-\mathrm{X}_{\mathrm{b}}\right)+1.8 \mathrm{X}_{\mathrm{b}}$ $\mathrm{X}_{\mathrm{b}}=0.256$ $\therefore \mathrm{X}_{\mathrm{t}}=1-\mathrm{X}_{\mathrm{b}}=0.744$

Hence, the correct answer is option: 'B'.

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