PROP 344: Molar Mass Determination by Freezing Point Depression in t-Butyl Alcohol

Post-Laboratory Questions

1. Obtain from your laboratory instructor the gram molar mass of your unknown. Calculate the percenterror for the gram molar mass of your unknown.

2. A student, following the procedure described in this module, used water as the solvent and encountered some interesting problems. Comment on the effect, if any, each of the following situations could have had on the experimental results.(1) The unknown, a white powder, failed to dissolve in the solvent.

Freezing Point depression is a solution phenomenon. The solute must dissolve in the solvent to create a colligative effect. In this solution, the experiment is not valid.

(2) The student returned to the laboratory instructor for a different solid unknown. This unknown dissolved, but bubbles were seen escaping from the solution almost immediately after the addition of the solid.

A chemical reaction occurred, as evidenced by the bubbles, so that the species in solution was changed. The experiment is not valid.

(3) As the student was setting up the apparatus to measure the freezing point of the unknown solution, the thermometer assembly rolled off the laboratory bench, and the thermometer broke. The student obtained a new thermometer and performed the experiment as instructed.

The Freezing point in both the solvent and of the solution must be measured with the same thermometer to eliminate any error caused by variation between the two different thermometers.

3. A student determined the Kf of t-butyl alcohol using tap water instead of distilled or deionized water.Describe the problems that might have been encountered. How would these problems affect the magnitude of Kf?

4. As a research chemist, you are interested in studying the extent and types of interactions in aqueous salt solutions. As part of this study, you weigh three samples of NaCI and dissolve each in 1.000 kgH2O. You then measure the freezing temperature of each solution and compare these temperatures to the freezing point of water. The data you collect are tabulated below. Explain the observed results.Predict and briefly explain the result you would expect for a solution made up of 29.22 g NaCI dissolved in 1.000 kg H2O.

g of NaCI per 1.000 Kg of H2O ∆Tf oC

5.845 0.348 0.585 0.0360 0.293 0.0182

There are several conclusions which can come from these data:

1. The gram molar mass of NaCI = 58.5 g mol-1. Therefore the solution contains 0.100, 0.001, and 0.005 mol of NaCI, respectively. The molality of the three solutions can be calculated as follows.

a. m = (5.85g ) 1mol NaCl

58.5g1.00Kg

= 0.100

b. m = (0.585g ) 1mol NaCl

58.5 g1.00Kg

= 0.01

c. m = (0.293g ) 1mol NaCl

58.5 g1.00Kg

= 0.005

The mc values calculated on the basis of ∆Tf are

a. mc = (0.348℃)

1.86℃kg /mol = 0.187

b. mc = (0.0360℃)1.86℃kg /mol

= 0.0194

c. mc = (0.182℃)

1.86℃kg /mol = 0.00978

A comparison of m and mc indicates that for NaCI solution mc ≅ 2M. This comparison implies that each unit of NaCI dissociates into two particles in water.

2. De data also implies that dissociation is more complete in dilute solutions that in concentrated solutions.a. m = 0.100

2m = 0.200; compare with mc = 0.187

b. m = 0.012m = 0.02; compare with mc = 0.0194

c. m = 0.0052m = 0.01; compare with mc = 0.00978

On the basis of these calculations, we would expect a solution of 29.22 g NaCI in 1 Kg H2O to have a freezing point depression of

∆Tf = (2M) (Tf), because mc ≅ 2M

m= (29.2g ) 1mol

58.5g1.00kg

= 0.499

so mc ≅ 2M = (2)(0.499) = 1.00 m

∆Tf = mcKf = (1.0)(1.86℃kg mol-1) = 1.86 ℃

But for a 0.5 m solution, the freezing point depression will be less than that, because mc ≅ 2M as the degree of dissociation decreases with increased concentration of the solute.