Calculate the mole fraction of benzene and naphthalene in the vapour phase when an ideal liquid solution is 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.

Explanation

`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`