Question

Assuming ideal solution behavior, what is the freezing point of a solution of dibromobenzene, C₆H₄Br₂, in 0.250 kg of benzene, if the solution boils at 83.5 °C?

a) 3.04 °C
b) 1.52 °C
c) 2.56 °C
d) 1.28 °C

Answer 1

The freezing point of the solution is approximately -26.73 °C.

Step-by-step explanation:

The freezing point depression is given by the equation:

ΔTf = Kf · m

Where ΔTf is the freezing point depression, Kf is the molal freezing point depression constant for benzene, and m is the molality of the solution.

First, we need to calculate the molality of the solution:

moles of dibromobenzene = mass / molar mass = (0.250 kg)(1000 g/kg) / (2(79.5 g/mol)) = 1.573 mol

molality = moles of solute / mass of solvent = 1.573 mol / 0.250 kg = 6.292 m

Now, we can calculate the freezing point depression:

ΔTf = (Kf)(m) = (5.12 °C/m)(6.292 m) = 32.23 °C

The freezing point of the solution is the freezing point of pure benzene minus the freezing point depression:

Freezing point of solution = 5.5 °C - 32.23 °C = -26.73 °C

Therefore, the freezing point of the solution is approximately -26.73 °C.