When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was x % of vapor pressure of pure solvent. What is x % ?

Step-by-step solution

Moles of solute, n_B = 0.25 Moles of solvent, n_A = 1 Mole fraction of solvent, X_A = n_A n_A + n_B = 1 1 + 0.25 = 1 1.25 = 0.8 According to Raoult's law, the vapor pressure of the solution is given by P_s = P^0 X_A P_s = 0.8 P^0 = 80 % of P^0 Thus, the vapor pressure of the solution is 80 % of the vapor pressure of the pure solvent. Answer: 80 %