Determine the freezing point of a solution that contains 0.8 mol C10H22 dissolved in 613.6 ml of benzene (density of benzene = 0.88 g/ml) A. –7.59°C B. –2.09°C C. +7.59°C D. +13.09°C E. Cannot be determined. - g

Answer B.

That option must be the right one. It depends on the decimals, I got -2.01°C and that is so closed.

Explanation:

We can calculate the freezing point of solution with the freezing point depression formula, of the colligative property as this:

Freezing T° of solution = - ( Kf . m . i - Freezing T° of pure solvent)

Kf for benzene: 5.07 °C/m

m = molality (mol/kg)

Mass of solvent (benzene) needs to be calculated with the density.

Benzene density = Benzene mass / Benzene volume

Benzene mass = Benzene density . Benzene volume

0.88 g/mL . 613.6 mL = 539.9 g

Let's convert the mass to kg → 539 .9 g . 1kg / 1000g = 0.5399 kg

m = 0.8 mol / 0.5399 kg → 1.48 m

As C₁₀H₂₂ is non electrolyte, i = 1

Freezing T° of pure solvent = 5.5°C

Let's replace:

Freezing T° of solution = - ( Kf . m . i - Freezing T° of pure solvent)

Freezing T° of solution = - (5.07°C/ m . 1.48 m . 1 - 5.5°C)

Freezing T° of solution = - 2.01 °C

wher is the rest of the question