Question

Freezing Point Depression Lab

Objective: To use freezing point to determine the molecular weight of an unknown substance.

In this lab, we used benzophene as the solvent, which has a freezing point of 48.1 degrees celsius and Kf value of 9.80 degrees C/m. The procedure consisted of measuring out 10 g benzophenone in a test tube and melting the substance in a warm water bath. When melted, we removed the test tube from the bath and allowed it to cool. We observed the freezing point. Next, we added half a gram of an unknown substance to the test tube and melted the soultion once more. We observed the freezing point.

My question is the following: How would the experiment be different if we had used water as the solvent? Explain what you would have to change to account for this difference and how you would do it.

Answer 1

In the freezing point depression experiment which is a colligative property, the freezing point is dependent on the molality of solution and the freezing point constant Kf.

Kf for benzophenone = 9.8 oC/m

Kf for water = 1.86 oC/m

If the experiment is done in water instead of benzophenone, The freezing point will now be more dependent on the interaction of solvent with the solute and the solute solute interactions which are more prominent in water due to its properties. Water is more polar than benzophenone and hence the freezing point observed would be different. The properties of water vary widely with temperature and is considered as non-ideal solvent for freezing point depression studies. Kf for benzophenone is much greater then water and we would see a difference in freezing point due to this as the depression in freezing point is also directly related to Kf value of solvent used for making the solution. We also see that the heat of fusion of water is different at different temperatures around the freezing point and the specific heat of water increases on supercooling but the specific heat of ice decreases on cooling. These properties make water a non-ideal solvent for accurate molar mass estimation by freezing point depression studies. To compensate this in the experiment we would have to run the reaction at higher temperatures.