A non-volatile solute A weighing 60 g when dissolved in 212 g of the solvent Xylene reduces its vapour pressure to 60 % . What is the Molar mass of A in g / mol ? [MM of xylene =106 ~g / mol ]

Step-by-step solution

According to Raoult's law for a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute: P^o - P_s P^o = n_A n_A + n_X Given that the vapour pressure is reduced to 60 % , we have P_s = 0.6 P^o . Therefore, P^o - 0.6 P^o P^o = 0.4 . The mole fraction of the solute x_A = n_A n_A + n_X = 0.4 . This implies n_X n_A = 1 - 0.4 0.4 = 0.6 0.4 = 1.5 . The number of moles of solvent Xylene is n_X = 212 g 106 g/mol = 2 mol . Substituting n_X into the ratio: 2 n_A = 1.5 n_A = 2