A solution is prepared by dissolving 53.5 grams of ammonium chloride, NH4Cl in 4000 grams of water ....?

Question

what is the freezing point of this solution?

Answer 1

First, calculate the molality of the solution:

1. Convert the mass of NH4Cl into moles by dividing by the molar mass.

2. Divide the moles by 4 kg solvent to give the molality.

Next, find the freezing point depression constant for water. Multiply the molality of the solution by the freezing point depression constant. For water, that will give you the freezing point of the solution.

Answer 2

NH4Cl MW = 53.5 so 563.5 g = 1 mole
1 mole per 4000 g = 0.25 mole /. 1000 grams water
Solution is 0.50 Molal in ions (NH4+ and Cl-)
fp depression = 1.86 x 0.5 = 0.93 deg C
fp = -0.93 deg C