Question

When 20.0 grams of an unknown compound are dissolved in 500. grams of benzene, the freezing point of the resulting solution is 3.77°C. The freezing point of pure benzene is 5.48°C, and the Kf for benzene is 5.12°C/m. What is the molecular weight of the unknown compound?

Answer 1

Answer: 120g/mol

Explanation:

The first step we are to take is to calculate the freezing point depression of the solution.

ΔT(f) = freezing point of pure solvent - freezing point of solution

ΔT(f) = 5.48 - 3.77

ΔT(f) = 1.71°C

Next we are to calculate the molal concentration of the solution using freezing point depression

ΔT(f) = K(f) * m

m = ΔT(f)/K(f)

m = 1.71/5.12

m = 0.333 molal

Now, we calculate the molecular weight of the unknown...

m = 0.333 mol = 0.333 mol X per kg of benzene

moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene

moles of X = 0.1665

molecular weight of X = 20g of X/0.1665

molecular weight of X = 120/mol

Answer 2

The molecular weight of the unknown compound is 120g/mol

Calculation of The molecular weight:

Here first determined the freezing point depression of the solution.

ΔT(f) = freezing point of pure solvent - freezing point of solution

ΔT(f) = 5.48 - 3.77

ΔT(f) = 1.71°C

Now the molar concentration should be

ΔT(f) = K(f) * m

m = ΔT(f)/K(f)

m = 1.71/5.12

m = 0.333 mol

Now, molecular weight should be

m = 0.333 mol = 0.333 mol X per kg of benzene

moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene

= 0.1665

Now

molecular weight of X = 20g of X/0.1665

= 120/mol

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