Molality and colligative property?

Question

For these questions.....please answer them step by step.

Q part1: A solution of a hydrocarbon dissolved in benzene freezes at -0.51(degree celcius). What is the molality of this solution?

Q part2: If the solution in (Q part1) was made by dissolving 4.367gram of the hydrocarbon in 21.35gram of benzene, how many mole of hydrocarbon were dissolved in the 21.35gram of benzene? + what is the molar mass of hydrocarbon?

please help me.. thanks =D

Answer 1

I am very surprised that a chem professor does not know benzene freezes at 5.5C.

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Answer 2

Use your molal equations. dT = i x Kf x m. So if dT = -0.51, i = 1, and Kf (should be a table in your chapter that gives it for benzene) = 5.12. So rearrange your equation to get:

dT / (i x Kf) = m or -0.51 / (1 x 5.12) = m

For the second part, once you solve and get m, remember that molality (m) = # moles/kg of solvent. The kg of the solvent is 0.02135, so

0.02135 x m = moles.

Remember that moles = grams / molecular weight, so m.w. = grams / moles. You know the grams, so plug it in:

4.367 grams / moles (from above) = m.w. of the material.