When 1 mole of benzene is mixed with 1 mole of toluene then what is the correct statement about composition of vapour of solution? (Given: Vapour pressure of benzene = 12.8 kPa and Vapour pressure of toluene = 3.85 kPa)

Higher percentage of benzene

Higher percentage of benzene because its vapour pressure is high.

mole fraction of benzene = \(\frac{1}{1+1}\) = \(\frac{1}{2}\) = 0.5

mole fraction of toluene = \(\frac{1}{1+1}\) = \(\frac{1}{2}\) = 0.5

p (benzene) = 12.8 x 0.5 = 6.4 kPa

p (toluene) = 3.85 x 0.5 = 1.925 kPa