Question

Calculate the freezing point of a solution made from 52.6 g of propane, c3H8, dissolved in 196.0 g of benzene, C6H6. The freezing point of benzene is 5.50 C and its Kf is 5.12 C/m.

Answer 1

Given: weight of solute (propane) = 52.6 g
weight of solvent (benzene) = 196 g = 0.196 kg
We know that, molecular weight of propane = 44.1 g/mol

∴ Molality of solution = $\frac{\text{weight of solute (g)}}{\text{Molecular weight X Weight of solvent (Kg)}}$
                                  =  $\frac{52.6}{44.1X.196}$ = 6.085 m

Now, Depression of freezing point = Kf m
where Kf = cryoscopic constant = 5.12 oC/m

∴  Depression of freezing point = 5.12 X 6.085
                                                  = 31.15 oC

Therefore, freezing point of solution = 5.50 - 31.15 = -25.65 oC