Question

Two solutions are prepared using the same solute:

Solution A: 0.27 g of the solute dissolves in 27.4 g of t-butanol.

Solution B: 0.23 g of the solute dissolves in 24.8 g of cyclohexane.

Which solution has the greatest freezing point change? Show calculations and Explain.

The kf (freezing point constant) of t-butanol = 9.1

The kf (freezing point constant) of cyclohexane= 20.0

Answer 1

Let the molecular weight of solute be M

Molality of Solution A=(Mass of solute/molecular wt of solute)/Mass of solvent in kg

                                                m=(0.27/M)/0.0274

                                                m=9.854/M

Depression in freezing point= k_f*m=9.1*9.854/M=89.67/M

Molality of Solution B=(Mass of solute/molecular wt of solute)/Mass of solvent in kg

                                                m=(0.23/M)/0.0248

                                                m=9.274/M

Depression in freezing point= k_f*m=20*9.274/M=185.5/M

As M is constant for both solution, solution B will have greatest freezing point change.