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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.
Solution
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
Now, no. of moles present in 80 g of benzene = `80/78 "mol"` = 1.026 mol
And, no. of moles present in 100 g of toluene = `100/92 "mol"` = 1.087 mol
∴ Mole fraction of benzene, xb = `(1.026)/(1.026 + 1.087)` = 0.486
And, mole fraction of toluene, xt = 1 − 0.486 = 0.514
It is given that the vapour pressure of pure benzene, `"p"_"b"^0` = 50.71 mm Hg
And, vapour pressure of pure toluene, `"p"_"t"^0` = 32.06 mm Hg
Therefore, the partial vapour pressure of benzene, `"p"_"b" = "x"_"b" xx "p"_"b"^0`
= 0.486 × 50.71
= 24.65 mm Hg
And, partial vapour pressure of toluene, `"p"_"t" = "x"_"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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