By the colligative the property of freezing point depression the molar mass of a solute can be determined. As an example, the freezing point, in degrees Celsius, of a pure solvent, t-butyl alcohol is measured in 5 different type of experiments with the following data: 25.10, 24.87, 25.03, 24.76, 24.94.
A solution is prepared dissolving 0.504 g of solute in 9.993gof t-butyl alcohol and the freezing point of the solution was measured in degrees Celsius, 5 times with the following data: 22.71, 22.52, 22.90, 22.64, and 22.83.
1. Determine the mean values and standard deviations of the freezing points for the pure solvent and the solution.
2. Determine the standard deviation for the difference among the 2 freezing points.
3. Determine the molar mass of the solute based on the freezing point depression given that the freezing point depression constant (Kf) for t-butyl alcohol is 8.09 *C/m.