Vapour pressure of CCl ₄ at 25^ C is 143 mm Hg. If 0.5 g of a non-volatile solute is dissolved in 100 cm ^3 of CCl ₄ . Find the vapour pressure of the solution. (Density of CCl ₄ = 1.58 g / cm ^3 and molecular weight of solute is 65 )

Step-by-step solution

Mass of solvent ( CCl₄ ), W₁ = Volume Density = 100 1.58 = 158 g Molar mass of CCl₄ , M₁ = 12 + 4(35.5) = 154 g/mol Number of moles of solvent, n₁ = W₁ M₁ = 158 154 1.026 Mass of solute, W₂ = 0.5 g Molar mass of solute, M₂ = 65 g/mol Number of moles of solute, n₂ = W₂ M₂ = 0.5 65 0.0077 According to Raoult's law, the relative lowering of vapour pressure is given by: P^0 - P_s P^0 = n₂ n₁ + n₂ 143 - P_s 143 = 0.0077 1.026 + 0.0077 143 - P_s 143 = 0.0077 1.0337 0.00745 143 - P_s = 143 0.00745 1.065 P_s = 143 - 1.065