The freezing point of a solution that contains *1.00 g* of an unknown compound, (A), dissolved in *10.0 g of benzene* is found to be 2.07 ^{o}C. The freezing point of pure benzene is 5.48 ^{o}C. The molal freezing point depression constant of benzene is 5.12 ^{o}C/molal. What is the molecular weight of the unknown compound?

- Step 1: Calculate the freezing point depression of the solution.

T_{f}= (Freezing point of pure solvent) - (Freezing point of solution)

(5.48^{o}C) - (2.07^{o}C) = 3.41^{o}C

- Step 2 : Calculate the molal concentration of the solution using the freezing
point depression.

T_{f}= (K_{f}) (m)

m = (3.41^{o}C) / (5.12^{o}C/molal)

*m = 0.666 molal*

**Step 3: Calculate the molecular weight of the unknown using the molal concentration.**

**m = 0.666 molal = 0.666 mol A / kg benzene**

**moles A = (0.66 mol A / kg benzene) (0.100 kg benzene) = 6.66 X 10**^{-3}**mol A**

**molecular weight of A = (1.00 g A) / (6.66 X 10**^{-3}**mol A)**

**molecular weigh of A = 150. g/mol**

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