Question

A 1.457 g of toluene (C₇H₈) was dissolved in 100.0 ml of cyclohexane. The density of cyclohexane is 0.779 g/ml. What was the change in the freezing point of the cyclohexane?

a) 1.2°C
b) 2.5°C
c) 3.8°C
d) 4.7°C

Answer 1

To calculate the change in freezing point of cyclohexane when toluene is dissolved in it, one needs the freezing point depression constant of cyclohexane, which is not provided. The molality of the solution can be calculated, but without the depression constant, the question cannot be accurately answered.

Step-by-step explanation:

The question asks about the change in the freezing point of cyclohexane when toluene is dissolved in it. This is a colligative property problem. Using the information given, we can calculate the molality (m) first: mass of solute (toluene) / (mass of solvent (cyclohexane) × molar mass of toluene). Cyclohexane's mass is 100.0 mL × 0.779 g/mL = 77.9 g.

Toluene's molar mass is 92.14 g/mol (as given by C₇H₈). Molality (m) = 1.457 g / (77.9 g × 92.14 g/mol) = 0.000162 mol/g. The freezing point depression constant (Kf) of cyclohexane is required to proceed further, and unfortunately, isn't provided in the question or the information given.

To find the freezing point depression (ΔTf), we use the formula ΔTf = Kf × m. Without the value for Kf, however, we cannot definitively solve for ΔTf. Based on sample solutions from similar problems, if we had a Kf value, we would then subtract the ΔTf from the pure solvent's freezing point to find the new freezing point of the solution.