Question

1. What is the relationship between the moles of solute and the mass of solute?
2. Rewrite the molality expression in terms of grams and molecular weight.

Background:

Properties that depend on the concentration of particles in solution are called colligative properties. The number of particles in a solvent can affect the freezing or boiling point of solvent. To express the effect of concentration on freezing point or boiling point, molality is used in place of molarity. Molality (m) is the number of moles of solute per 1 kilogram of solvent. moles of solute molality (m) Kg of solvent The presence of a solute affects the boiling point of a solvent by lowering the vapor pressure of that solvent. The magnitude of this reduction in vapor pressure depends upon the concentration of solute particles present in the solvent. Since the freezing point of a liquid depends on its vapor pressure, introduction of a solute in a solvent will alter the boiling point of the solution from that of the pure solvent It is often necessary to determine the freezing point of a solvent or solution using cooling curves which are plots of time versus temperature. Generally, these plots show a steep decrease in temperature over time, followed by a horizontal plateau. The point where these two lines meet corresponds to the freezing point of the substance being measured. Figure 1 shows an example of a cooling curve. Figure 1: Sample Cooling Curve for Benzophenone Cooling Curve 90 80 70 60 50 40 30 20 10 50 100 150 200 250 300 Time (s) Relevant equations related to freezing point depression AT Tsolvent) T(solution) AT K m where Tis freezing point (Recall that the freezing point is the same as the melting point.), K; is the freezing point constant for the solvent, and m is the molality of the solute. The solvent used in the experiment is benzophenone. The expected freezing point (T{solvent)) for benzophenone is 48.1°C. Even though it is printed here, you should experimentally determine the freezing point. The K for benzophenone is 9.80 °C/m. You will be given an unknown solid. You will used the colligative property to calculate the molar mass of this unknown. Example calculations are given in your textbook. Temperature (deg C

Answer 1

moles of solute = (mass of solute / molar mass of solute); where mass is measured in grams and molar mass (defined as the mass of one mole of a substance in grams) is measured in g/mol.

molality=((mass of solute)/(molecular weight of solute*kg of solvent)