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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 $]$
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The correct answer is:
$0.744$
$1.013=0.742 \mathrm{X}_{\mathrm{t}}+1.8 \mathrm{X}_{\mathrm{b}}$
$1.013=0.742\left(1-\mathrm{X}_{\mathrm{b}}\right)+1.8 \mathrm{X}_{\mathrm{b}}$
$\mathrm{X}_{\mathrm{b}}=0.256$
$\therefore \mathrm{X}_{\mathrm{t}}=1-\mathrm{X}_{\mathrm{b}}=0.744$
$1.013=0.742\left(1-\mathrm{X}_{\mathrm{b}}\right)+1.8 \mathrm{X}_{\mathrm{b}}$
$\mathrm{X}_{\mathrm{b}}=0.256$
$\therefore \mathrm{X}_{\mathrm{t}}=1-\mathrm{X}_{\mathrm{b}}=0.744$
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