A solution of a non-volatile solute in a hypothetical solvent X has a boiling point elevation of 2.5^ C . The ebullioscopic constant ( K_b ) of solvent X is 2.5 K kg mol ⁻¹ and its molar mass is 50 g mol ⁻¹ . If the vapour pressure of pure solvent X at a given temperature is 420 mm Hg , what is the vapour pressure of the solution at the same temperature?

Step-by-step solution

First, calculate the molality ( m ) of the solution from the boiling point elevation: T_b = K_b m 2.5 = 2.5 m m = 1 mol kg ⁻¹ This means 1 mol of solute is dissolved in 1 kg ( 1000 g ) of solvent X. Moles of solvent X in 1000 g : n_ solvent = 1000 50 = 20 mol Mole fraction of solute ( X_ solute ): X_ solute = n_ solute n_ solute + n_ solvent = 1 1 + 20 = 1 21 According to Raoult's Law, the relative lowering of vapour pressure is: P^0 - P_s P^0 = X_ solute 420 - P_s 420 = 1 21 420 - P_s = 420 1 21 = 20 mm Hg P_s = 4