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Question
Benzene and toluene form 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.
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Solution
Given: Vapour pressure of pure benzene `(p_b^0)` = 50.71 mm Hg
Vapour pressure of pure toluene `(p_t^0)` = 32.06 mm Hg
Molar mass of benzene (C6H6) = 6 × 12 + 6 × 1
= 78 g mol−1
Molar mass of toluene (C6H5CH3) = 7 × 12 + 8 × 1
= 92 g mol−1
Number of moles present in 80 g of benzene = `80/78` mol
= 1.026 mol
Number of moles present in 100 g of toluene = `100/92` mol
= 1.087 mol
Mole fraction of benzene (χb) = `(1.026)/(1.026 + 1.087)`
= 0.486
Mole fraction of toluene (χ) = 1 − 0.486
= 0.514
Partial vapour pressure of benzene (pb) = `chi_b xx p_b^0`
= 0.486 × 50.71
= 24.65 mm Hg
Partial vapour pressure of toluene (pt) = `chi_t xx p_t^0`
= 0.514 × 32.06
= 16.48 mm Hg
As a result, the mole fraction of benzene in the vapour phase is as follows:
`p_b/(p_b + p_t)`
= `24.65/(24.65 + 16.48)`
= `24.65/41.13`
= 0.599
= 0.6
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