Benzene and toluene form an 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 the vapour phase if $80 \, g$ of benzene is mixed with $100 \, g$ of toluene.

(0.6) Molar mass of benzene $(C_6H_6) = 78 \, g \, mol^{-1}$.
Molar mass of toluene $(C_6H_5CH_3) = 92 \, g \, mol^{-1}$.
Moles of benzene $(n_b) = \frac{80}{78} = 1.026 \, mol$.
Moles of toluene $(n_t) = \frac{100}{92} = 1.087 \, mol$.
Mole fraction of benzene $(x_b) = \frac{1.026}{1.026 + 1.087} = 0.486$.
Mole fraction of toluene $(x_t) = 1 - 0.486 = 0.514$.
Partial pressure of benzene $(p_b) = x_b \times p_b^o = 0.486 \times 50.71 = 24.645 \, mm \, Hg$.
Partial pressure of toluene $(p_t) = x_t \times p_t^o = 0.514 \times 32.06 = 16.479 \, mm \, Hg$.
Mole fraction of benzene in vapour phase $(y_b) = \frac{p_b}{p_b + p_t} = \frac{24.645}{24.645 + 16.479} = \frac{24.645}{41.124} = 0.599 \approx 0.6$.