A hypothetical solvent X has a molar mass of 50 g mol ⁻¹ and a normal boiling point of 80^ C . When a non-volatile solute is dissolved in X , the boiling point of the resulting solution is observed to be 90^ C . Given that the boiling point elevation constant ( K_b ) of solvent X is 2.0 K kg mol ⁻¹ , the vapour pressure of the solution at 80^ C is _____ mm Hg .

Step-by-step solution

Since the normal boiling point of solvent X is 80^ C , its vapour pressure in the pure state at this temperature is equal to atmospheric pressure: p^ = 1 atm = 760 mm Hg The elevation in boiling point is: T_b = 90^ C - 80^ C = 10^ C Using the formula for boiling point elevation: T_b = i K_b m 10 = i 2.0 m i m = 5 mol kg ⁻¹ This means there are 5 effective moles of solute particles in 1 kg ( 1000 g ) of solvent X . Moles of solvent X in 1 kg : n_ solvent = 1000 50 = 20 moles Using Raoult's Law, the vapour pressure o