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.

Explanation

Molar mass of benzene `(C_6H_6) = 78 g "mol"^(-1)`
Molar mass of toluene `(C_6H_3CH_3) = 92 g "mol"^(-1)`
` therefore ` Number of moles in 80 g of benzene ` = (80g)/(78 g"mol"^(-1)) = 1.026` mole
Number of moles in 100 g toluence ` = (100g)/(92 g "mol"^(-1)) = 1.087` mole
Mole fraction of benzene ` = (1.026)/(1.026 + 1.087) = (1.026)/(2.113) = 0.486`
Mole fraction of toluene = 1 - 0.486 = 0.514
`p_("benzene")^0 = 50.71 mm, p_("toluence")^0 = 32.06 mm`
`p_("benzene") = x_("benzene") xx p_("benzne")^0 = 0.486 xx 50.71 mm= 24.65 mm`
`p_("toluence") = x_("toluence") xx p_("toulence")^0 = 0.514 xx 32.06mm = 16.48 mm`
Mole fraction of benzene in the vapour phase
` = (p_("benzene"))/(p_("benzene") + p_("toluence")) = (24.65)/(24.65 + 16.48) = (24.65)/(41.13) = 0.60`