Benzene and toluene form ideal solution occur the entire range of composition The vapour pressure of pure benzene and toluene at $300 K$ are $50 : 71$ mm $H g$ and $32.06$ mm $H g$ respectively calculate the mole fraction of benzene is vapour phase if $80 g$ of benzene is mixed [ 100 g of toluene ].

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Solution

Number of moles of benzene= $\frac{80 g}{78 g m o l^{- 1}} = 1.026 m o l e$
Number of moles of toulene= $\frac{100 g}{92 g m o l^{- 1}} = 1.087 m o l e$
Mole fraction of benzene= $\frac{1.026}{1.026 + 1.087} = 0.486$
Mole fraction of toulene= $1 - 0.486 = 0.514$
$P_{b e n z e n e} = x_{b e n z e n e} \times P_{b e n z e n e}^{o} = 0.486 \times 50.71 = 24.65 m m$
$P_{t o u l e n e} = x_{t o u l e n e} \times P_{t o u l e n e}^{o} = 0.514 \times 32.06 = 16.48 m m$
Mole fraction of benzene in vapour phase= $\frac{P_{b e n z e n e}}{P_{b e n z e n e} + P_{t o u l e n e}}$
$= \frac{24.65}{24.65 + 16.48} = 0.60$

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Benzene and toluene form ideal solution occur the entire range of composition The vapour pressure of pure benzene and toluene at 300K are 50:71 mm Hg and 32.06 mm Hg respectively calculate the mole fraction of benzene is vapour phase if 80g of benzene is mixed [ 100 g of toluene ].