A student determined the molar mass of an unknown non-dissociating liquid by the

Question

A student determined the molar mass of an unknown non-dissociating liquid by the method described in this experiment. She found that the equilibrium temperature of a mixture of ice and pure water was indicated to be +0.4 degree C on her thermometer. When she added 9.9 g of her sample to the mixture, the temperature, after thorough stirring, fell to -3.7 degree C. She then poured off the solution through a screen into a beaker. The mass of the solution was 84.2 g. a. What was the freezing point depression? b. What was the molality of the unknown liquid? c. What mass of unknown liquid was in the decanted solution? d. What mass of water was in the decanted solution? e. How much unknown liquid would there be in a solution containing 1 kg of water, with her unknown liquid at the same concentration as she had in her experiment? f. Based on these data, what value did she calculate for the molar mass of her unknown liquid, assuming she carried out the calculation correctly?

Explanation / Answer

(a)

Freezing point depression = T0 - Tf = 0.4 - ( - 3.7) = 4.1 0C

(b)

We know that, Kf of water = 1.86 0C/m

Formula,

Depression in freezing point = Kf * m

4.1 = 1.86 * m

m = molality = 2.20 mol/kg

(c)

Mass of unknown liquid = 9.9 g.

(d)

Mass of water in solution = mass of solution - mass of unknown liquid = 84.2 - 9.9 = 74.3 g.

(e)

In 84.2 g. of solution 74.3 g. of water and 9.9 g. of unknow liquid is present.

then, 1 kg. (1000 g.) of water has 1000 * 9.9 / 74.3 = 133.2 g. of unkown liqiud.

(f)

molality = moles of solute / kg. of solvent

2.20 = n / 1

n = 2.20 moles

But number of moles = mass / molar mass

2.20 = 133.2 / molar mass

Therefore, molar mass of unknown liquid = 133.2 / 2.20 = 60.6 g./mol

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