The boiling point of water(100°C) becomes 100.25°C, If 3 grams of a nonvolatile solute is dissolved in 200ml of water. The molecular weight of solute is (Kbfor water is 0.6 K Kg / mole)

First boiling point of water = 100°CFinal  boiling point of water = 100.52°CWs = 3g , W0=200g , Kb=0.6K Kg mol-1 ∆Tb=100.52-100 =0.52°C ∆Tb= Kbm=KbWsMs×1000Wo Ms =Kb×Ws×1000∆Tb×Wo=0.6×3×10000.52×200=1800104=17.3 gmol-1 Where , Ws=weight of solute ,  W0=weight of solvent