100 g of liquid A (molar mass 140 g mol − 1 ) was dissolved in 1000 g of liquid B (molar mass 180 g mol − 1 ). The vapour pressure of pure liquid B was found to be 500 torr . Calculate vapour pressure of A (in torr) the solution if the total vapour pressure of the solution is 475 torr .

Step-by-step solution

Step I : Calculation of vapour pressure of pure liquid A (p° A ) ​Number of moles of liquid A n A = W A M A = (100 g) 140 g mol - 1 = 0.7143 mol Number of moles of liquid B n B = W B M B = (1000 g) 180 g mol - 1 = 5.5556 mol Mole fraction of A x A = n A n A + n B = (0.7143 mol) (0.7143 + 5.5556) mol = 0.7143 6.2699 = 0.1139 Mole fraction of B(x B ) = 1 − 0.1139 = 0.8861 Vapour pressure of pure liquid B, B(p B o ) = 500 torr Total vapour pressure of solution (p) = 475 torr According to the Raoult's law p = p A o x A