Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at $300 K$ are $50.71 m m H g$ and $32.06 m m H g$ respectively. Calculate the mole fraction of benzene in vapour phase if $80 g$ of benzene is mixed with $100 g$ of toluene.

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Solution

Number of moles is the ratio of mass to molar mass.

The molar masses of benzene and toluene are $78$ g/mol and $92$ g/mol respectively.

Number of moles of benzene $= \frac{80}{78} = 1.026$

Number of moles of toluene $= \frac{100}{92} = 1.087$

Mole fraction of benzene, $X_{B} = \frac{1.026}{1.026 + 1.087} = 0.486$

Mole fraction of toluene, $X_{T} = 1 - 0.486 = 0.514$

$P_{B} = P_{B}^{o} \times X_{B} = 50.71 \times 0.486 = 24.65$ mm Hg

$P_{T} = P_{T}^{o} X_{T} = 32.06 \times 0.514 = 16.48$ mm of Hg

Total vapour pressure $= 24.65 + 16.48 = 41.13$ mm Hg

Mole fraction of benzene in vapour phase is as follows:
$Y_{B} = \frac{24.65}{41.13} = 0.60$

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Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.