An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Given that, PS=1.004barPS=1.004bar W2=2gW2=2g Mass of solution = 100gMass of solution = 100g W1=100−2=98gW1=100−2=98g M1=18gmol−1M1=18gmol−1 We know that VP of pure water, P0=1atm=1.013barP0=1atm=1.013bar By Raoult’s law; 1.013−1.0041.013=2×18M2×981.013−1.0041.013=2×18M2×98 Thus, the molar mass of the solute, M2=41.35gmol−1M2=41.35gmol−1
