Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at T K are 50 mmHg and 40 mmHg , respectively. What is the mole fraction of toluene in vapour phase when 117 g of benzene is mixed with 46 g of toluene? (Molar mass of benzene and toluene are 78 and 92 g mol - 1 , respectively)

Step-by-step solution

Given, Vapour pressure of benzene, P B ° = 50   mmHg Vapour pressure of toluene, P T ° = 40   mmHg Molar mass of benzene = 78   g   mol - 1 Molar mass of toluene = 92   g   mol - 1 Moles of benzene = 117 78 = 1 . 5 Moles of toluene = 46 90 = 0 . 51 Mole fraction of benzene = 1 . 5 1 . 5 + 0 . 511 = 0 . 75 Mole fraction of toluene = 0 . 511 1 . 5 + 0 . 511 = 0 . 25 P B = P B ° X B = 50 × 1 . 5 = 75   mmHg P T = P T ° X T = 40 × 0 . 511 = 20 . 44