A student determines the molar mass of acetone, CH3COCH3, by the method used in

Question

A student determines the molar mass of acetone, CH3COCH3, by the method used in this experiment.  She found that the equilibrium temperature of a mixture of ice and water was 1.0 degrees C on her thermometer.  When she added 11.1 g of her sample to the mixture, the temperature, after thorough stirring, fell to -3.0 degrees C.  She then poured off the solution through a screen into a beaker.  The mass of the solution was 90.4 g.

a) What was the freezing point depression?


b) What was the molality of acetone? I need help especially with this one.
.

c) How much acetone was in the decanted solution?


d) How much water is in the decanted solution?


e) How much acetone would there be in a solution containing 1 kg of water and acetone at the same concentration as she had in her experiment?


f) What did she find to be the molar mass of acetone, assuming she made the calculation properly?

Explanation / Answer

a. It is clear that the initial temperature of the mixture of Ice and Water was 1 degree C. But after adding acetone, it was fallen to -3 degree C. So the depression in freezing point is = initial temperature - final temperature
so freezing point depression = 1 - (-3) = 4
so your answer for the first part is absolutely correct. :-)

b. Molality is = moles of solute/ weight of the solvent in Kg.
Acetone is the solute and water is the solvent. weight of acetone is given = 11.1 g
weight of the solution is = 90.4 g
we know that solution = solute + solvent
so the weight of the solvent that is water = 90.4 - 11.1 = 79.3 g
moles of solute = weight of the solute in grams/molar mass of the solute
molar mass of acetone = 58 g/mole
moles of acetone = 11.1 / 58 = 0.19 moles
molality = 0.19 x 1000 / 79.3 = 2.39 = 2.4 m

c and d parts are correct.

e. In the experiment, 79.3 g water contained 11.1 g of acetone.
1 g water will contain 11.1 / 79.3 g acetone.
and hence 1 Kg that 1000 g water will contain 11.1 x 1000 / 79.3 g of acetone
= 139.97 g acetone.

f. To calculate the molar mass we need to take the sum of the atomic masses of all the atoms present in acetone.
CH3COCH3 = 12+1x3 + 12 + 16 + 12 + 1 x 3 = 58

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