Calculate the freezing point of a solution made from 22.0 g of octane (C₈H₁₈)dissolved in 148.0 g of benzene.Benzene freezes at 5.50°C and its K_f value is 5.12°C/m. A)-1.16°C B)0.98°C C)5.49°C D)6.66°C E)12.2°C

Question

Quizplus · Accepted Answer

First, calculate the molality of the solution:

molality = moles of solute / mass of solvent in kg

The molar mass of octane is 114.23 g/mol.

moles of octane = 22.0 g / 114.23 g/mol = 0.1927 mol

mass of solvent in kg = 148.0 g / 1000 = 0.148 kg

molality = 0.1927 mol / 0.148 kg = 1.3013 m

Next, calculate the change in freezing point using the formula:

ΔTf = Kf x molality

ΔTf = 5.12°C/m x 1.3013 m = 6.6664°C

Finally, calculate the freezing point of the solution:

freezing point = freezing point of pure solvent - ΔTf

freezing point = 5.50°C - 6.6664°C = -1.1664°C

Rounding to two decimal places, the freezing point of the solution is -1.16°C. Therefore, the best choice is A.

Calculate the Freezing Point of a Solution Made from 22