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.

Benzene and toluene form ideal solution over the entire range of compo

1.	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.
Answer» <P> Solution :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`