1. A student determines the molar mass of acetone, CHCOCH, by the method used in

Question

1. A student determines the molar mass of acetone, CHCOCH, by the method used in this experiment. She found that the equilibrium temperature of a mixture of ice and water was 1.0°C on her thermometer When she added 11.1 g of her sample to the mixture, the temperature, after thorough stirring, fell to -3.00. She then poured off the solution through a screen into a beaker. The mass of the solution was a. What was the Areezing point depression? oc b. 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? 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? e. -g acetone f. What did she find to be the molar mass of acetone, assuming she made the calculation properly?

Explanation / Answer

Solution:

1)

freezing point depression =; dTf = freezing point of pure water - freezing point of solution

freezing point of pure water = 0.0C

freezing point of solution = -3.0 C

therefore;

dTf = 0 - (-3)

dTf = 3.0C

Thus the freezing point depression is 3 C

2)

freezing point depression;dTf = Kf x m

here ;

Kf for water = 1.86 C kg /mol

And freezing point depression= 3.0C

m = molaity

Then;

3 = 1.86 x m

m = 1.613

The molality of the unknown solution is 1.613 m

3)

11.1 g of unknown liquid was in the decanted solution

4)

Given that the mass of solution = 90.4 g

Mass of solution = mass of solute + mass of solvent

Here solvent is water. Thus

amount of water = 90.4g - 11.1g

amount of water = 79.3 g

so there is 79.3 g of water in the decanted solution .

5)

We know that molality is the number of moles in per kg of solvent then;

Here mass of water = 79.3 g

79.3 g *1.0 kg /1000 g = 0.0793 kg

molality = moles of liquid / mass of water (kg)

1.613 = moles of liquid / 0.0793

moles of liquid = 0.128 moles

and number of moles = amount in g / molar mass

then;

molar mass = amount in g / number of moles

= 11.1 g / 0.128 moles

=86.72 g / moles

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