Question

Colligative Properties: Freezing Point Depression

Lauris acid is added to a test tube and heated. The test tube is placed in water to freeze the lauric acid. After measuring the temperature, an unknown is added. The lauric acid and unknown are heated to a liquid and placed in cold water(Part B). In doing part B of this experiment, Group 1 removes the thermometer and stirring wire to add the unknown, and wipes them off, losing 0.25 g of lauric acid. If Group 2 adds the same unknown without losing lauric acid, which group should measure a larger freezing point depression? For Group 1, what percent error will this introduce if they had originally weighed 10 g of lauric acid and continued to use this value in their calculations?

I believe for the first question, Group 1 would measure a larger freezing point depression. However, I don't know how to calculate the percent error for the second question.

Answer 1

Since freezing point depression is calculated by the formula:

$\Delta$ T = K_f*m

Here, m is the molality of solution

So, m = moles of solute / mass of solvent in kgs

Since moles of solute added are same for both groups, the only difference is mass of solvent. Group A has a lesser mass of solvent because of the loss encountered, as a result the molality becomes larger than it actually was for group A. Thus group A registers a larger depression in freezing point.

For lauric acid, K_f = 3.9 'C/m

Assume that 1 mole of unknown is added.

So, Actual $\Delta$ T for group A because of the loss = 3.9*( 1 / ( (10-0.25)*10-3) ) = 400 'C

$\Delta$ T as reported by group A ( being un aware of the loss, and hence the error ) = 3.9*( 1 / ( 10*10-3) ) = 390 'C

So, error % = Difference / Actual value * 100 = 10/400 * 100 = 2.5%