Advance Study Assignment: Determination of Molar Mass by Depression of the Freez

Question

Advance Study Assignment: Determination of Molar Mass by Depression of the Freezing Point A student determines the molar mass of methanol, CH,OH, by the method used in this experiment. She found that the equilibrium temperature of a mixture of ice and pure water was 0.4°C on her thermometer When she added 10.0g of her sample to the mixture, the temperature, after thorough stirring, fell to -5.4. She then poured off the solution through a screen into a beaker. The mass of the solution was 101.8 g. I. a. What was the freezing point depression? b. What was the molality of the methanol? c. How much methanol was in the decanted solution? d. How much water was in the decanted solution? How much methanol would there be in a solution containing 1 kg of water, with methanol at the same concentration as she had in her experiment? e. g methanol f. What did she find to be the molar mass of methanol, assuming she made the calculation properly?

Explanation / Answer

a.

Given that;

initial temperature = 0.4C

final temperature = -5.4 c

depression in freezing point is = initial temperature - final temperature
so freezing point depression = 04 - (-5.4) = 5.8

b.

Molality = moles of solute/ weight of the solvent in Kg.
CH3OH, methanol is the solute and water is the solvent. weight of CH3OH is given = 10.0 g
weight of the solution is = 101.8 g
we know that solution = solute + solvent
so the weight of the solvent that is water = 101.8 - 10 = 91.8 g
moles of solute = weight of the solute in grams/molar mass of the solute
molar mass of CH3OH = 32.04 g/mole
moles of CH3OH= 10 / 32.04= 0.312moles
molality = 0.312 x 1000 / 91.8 = 3.398 = 3.4 m

c) 10 g

d) 91.8 g
weight of the solution is = 101.8 g
we know that solution = solute + solvent
so the weight of the solvent that is water = 101.8 - 10 = 91.8 g

e. In the experiment, 91.8 g water contained 10 g of CH3OH.
1 g water will contain 10 / 91.8 g CH3OH.
and hence 1 Kg that 1000 g water will contain 10 x 1000 / 91.8 g of CH3OH= 109 g CH3OH

f. To calculate the molar mass:

10.0 grams CH3OH/ 91.8 g H2O x 1000 g / kg x 1 kg H2O/ 3.12 moles CH3OH = 34.9 grams methanol/ 1mole methanol

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