Given that the freezing point of benzene is 5.48^ C and its K _ f value is 5.12 ^ C / m . What would be the freezing point of a solution of 20 ~g of propane in 400 ~g of benzene?

Step-by-step solution

The molar mass of propane ( C₃H₈ ) is 3 12 + 8 1 = 44 g/mol . The number of moles of propane is n = 20 g 44 g/mol = 5 11 mol 0.4545 mol . The mass of the solvent (benzene) is 400 g = 0.4 kg . The molality ( m ) of the solution is m = moles of solute mass of solvent in kg = 5/11 0.4 = 5 4.4 = 50 44 = 25 22 m 1.136 m . The depression in freezing point is given by T_ f = K_ f m . Substituting the values, T_ f = 5.12 25 22 = 128 22 5.818^ C . The freezing point of the solution is T_ f = T_ f ^ - T_ f = 5.48^ C - 5.818^