Solution Preparation

To prepare a 5% solution of CaCl2, we need to dissolve 5 grams of CaCl2 in 100 grams of solution. Therefore, to prepare a 6g NaCl solution with 5% CaCl2 by weight, we need to dissolve 0.3g CaCl2 (5% of 6g) in 5.7g water and then dissolve 6g NaCl in this solution. Finally, we need to add enough water to make the total volume one litre.

Freezing Point Depression

Freezing point depression is a colligative property of solutions that is defined as the difference between the freezing point of a pure solvent and the freezing point of a solution. The freezing point depression is directly proportional to the molality of the solution. The molality of a solution is defined as the number of moles of solute per kilogram of solvent.

The freezing point depression can be calculated using the following equation:

ΔTf = Kf * m

Where ΔTf is the freezing point depression, Kf is the freezing point depression constant of the solvent, and m is the molality of the solution.

For water, the freezing point depression constant is 1.86 °C/m. Therefore, for a 6g NaCl solution, the molality can be calculated as follows:

Molar mass of NaCl = 58.44 g/mol

Number of moles of NaCl = mass / molar mass = 6 / 58.44 = 0.1027 mol

Mass of water = 1000 – 6 = 994g

Molality of NaCl solution = 0.1027 / 0.994 = 0.103 mol/kg

Therefore, the freezing point depression can be calculated as follows:

ΔTf = 1.86 * 0.103 = 0.1916 °C

Therefore, the depression in freezing point of the 6g NaCl (5% by weight of CaCl2) solution in one litre of water is 0.1916 °C.