Abhiraj Pathak
Last Activity: 6 Years ago
Delta Tf = Tfo - Tf = Kf*m = Tf* w2/(M2*w1)where Tfo is freezing point of water, Tf is freezing point of solution. Kf is crycoscopic constant, w2 mass of solute, M2 molar mass of solute and w1 mass of solvent.Tfo = 0 degC, Tf = -0.465, w2 = 6.35g, w1 = 0.5 kg, Kf = 1.86 K/molal.The value of Kf is given in problem most of the time, if not remember Kf for water is 1.86Substituting the values, we get, M2 = 50.8 g/mol