Experiment 19 Advance Study Assignment Determination of Molar Mass by Depression

Question

Experiment 19 Advance Study Assignment Determination of Molar Mass by Depression of the Freezing Point I. A student determines the molar mass of acetone, CH,coCH, by the method used in this experiment found that the equilibrium temperature of a mixture of ice and water was 1.0°C on her thermometer, hen she added 1 g of her sample to the mixture, the temperature, after thorough fell -3.0°C. then poured off the solution through a screen into a beaker The mass of the solution was he 90.4 g. a. What was the freezing point depression? What was the molality of the acetone? c. How much acetone was in the decanted solution? d. How much water was in the decanted solution? e. How much acetone would there be in a solution containing 1 kg of water and acetone at the same con- centration as she had in her experiment? g acetone f what did she find to be the molar mass of acetone, assuming she made the calculation properly?

Explanation / Answer

a)

Freezing point depression was 1-(-3) = 4°C

b)

To determine molality of acetone we will use formula m = T/Kf

Here, T is temperature difference and Kf is molal freezing point depression constant of the solvent (1.86 °C/m)

m = 3/1.86 = 1.6 mol/kg

c)

11.1 g acetone was in the decanted solution.

d)

90.4 – 11.1 = 79.3

79.3 g water was in decanted solution

e)

We have 1.6 mol acetone/kg.

1 mole acetone = 58.07 g

1.6 mole acetone = 92.92 g/kg

f)

mass unknown g /kg water = MM * 1.6 mol/kg

MM = 11.1/ 0.079*1.6 = 87.81 g

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