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Freezing Point of a Solution of an Unknown Substance

Add 2g of the unknown solid #1 and 10 mL of water to the test tube. the constant temperature bath is still set to dry ice. Drag and drop the constant temperature bath onto the test tube containing the solution.
Repeat the measurement of the freezing point with unknown solid #2 using a new clean test tube. As with solid #1 use 2g of solute and 10 mL of water.

1. Record the measured freezing point of pure water. 0 C
2. Record the freezing points and freezing point depressions, Td, of:
(a) unknown #1: -2.06 C
(b) unknown #2: -4.40 C

3. For each unknown solution, record the mass of water, the mass of the unknown compound and calculate the molality.
(a) unknown #1
(b) unknown #2

4. Calculate the molecular weight, MW, in g/mole for each unknown.
(a) unknown #1
(b) unknown #2

5. If you are told that one of the unknowns is glucose, C6HO6, with a molecular weight of 180.16 g/mole, which would you say it is?

6. What effect would each of the following have on the calculated molecular weight of an unknown? Would the calculated value be higher or lower than the actual value? Explain your answers.
(a) Some of the unknown does not dissolve.
(b) The thermometer reads 0.63°C higher than it should over the whole temperature range.
(c) The test tube is not dry (and has water drops inside) before the solutions are made up.

Solution Preview

1. Record the measured freezing point of pure water. 0 C (ANSWERED)
2. Record the freezing points and freezing point depressions, Td, of:
(a) unknown #1: -2.06 C (ANSWERED)
(b) unknown #2: -4.40 C (ANSWERED)

3. For each unknown solution, record the mass of water, the mass of the unknown compound and calculate the molality.
(a) unknown #1

Td = k(f) m

Td = change in freezing point = T(initial) - T(final) = 0 C - -2.06 C = 2.06 C
k(f) = molal freezing constant for water = 1.86 C/kg/mol
m = molality of the solution.

From this calculation we will find molality. The k(f) value is a constant value for water and can be found in any Chemistry textbook (or from http://www.cliffsnotes.com/study_guide/Freezing-and-Boiling-Points.topicArticleId-21729,articleId-21699.html). The change in temperature is always set up to be a positive value, because we are strictly looking at a DIFFERENCE in temperature.

Td = k(f) m
m = Td/k(f) = 2.06 C/(1.86 C/kg/mol) = 1.11 mol/kg

Because we divided by kg/mol, we end up with mol/kg as units. The C cancels.

The molality of unknown #1 is 1.11 mol/kg

(b) unknown #2

Td = k(f) m

Td = change in freezing point = T(initial) - T(final) = ...

Solution Summary

Freezing points are used to determine the molecular weight of solutes. A colligative property (freezing point depression) is employed in this calculation. All work is shown and explained.

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