Calculate the mole fraction of benzene and naphthalene in the vapour phase when an ideal liquid solutionis formed by mixing 128 g of naphthalene with 39 g of benzene. It is given that the vapour pressure of pure benzene is 50.71 mm Hg and the vapour pressure of pure naphthalene is 32.06 mm Hg at 300 K.

Calculate the mole fraction of benzene and naphthalene in the vapour p

1.	Calculate the mole fraction of benzene and naphthalene in the vapour phase when an ideal liquid solutionis formed by mixing 128 g of naphthalene with 39 g of benzene. It is given that the vapour pressure of pure benzene is 50.71 mm Hg and the vapour pressure of pure naphthalene is 32.06 mm Hg at 300 K.
Answer» Solution :`P_("PURE BENZENE") ^(@) = 50.71 mm Hg` ` P _("nephtalene") ^(@)=32.06 mm Hg` Number of moles of benzene `=(39)/(78) =0.5 mol` Number of moles of naphthalene `= (128)/(128) =1 `mol MOLE fraction of benene `= (0.5)/(1.5) =0.33` Mole fraction of naphthalene `=1 -0.33 = 0.67` Partial vapour pressure of benzene `= P _(benzene")^(@) XX` Mole fraction of benzene `=50.71 xx 0.33 =16.73 `mm Hg Partial vapour pressure of naphthalene `= 32. 06 xx 0.67 = 21. 48 mm Hg` mole fraction of benzene in vapour PHASE `= (16.73)/(16.73 + 21. 48) = (16.73)/(38.21) = 0.44` Mole fraction of naphthalene in vapour phase `=1 -0.44 =0.56`