A solution is prepared by dissolving 53.5 grams of ammonium chloride, NH₄Cl, in 4000.0 grams of water. What is the freezing point of the solution? (The freezing point depression constant for water is 1.86°C/mole solute in 1000 g of water.) A) -0.37°C B) -0.47°C C) -0.93°C D) -1.86°C

Question

Quizplus · Accepted Answer

Firstly, we need to determine the molality of the solution, which is given by:
molality = (moles of solute) / (mass of solvent in kg)
The molar mass of ammonium chloride is 53.5 g/mol, so we have:
moles of NH4Cl = mass / molar mass = 53.5 g / 53.5 g/mol = 1 mol
mass of water = 4000.0 g = 4.0000 kg
molality = 1 mol / 4.0000 kg = 0.2500 m

Next, we can use the freezing point depression equation:
ΔTf = Kf x molality
where ΔTf is the change in freezing point, Kf is the freezing point depression constant (1.86°C/m for water), and molality is the molality of the solution.

ΔTf = (1.86°C/m)(0.2500 m) = 0.465°C
Since the freezing point is lowered by 0.465°C, the freezing point of the solution is:
freezing point = 0°C - 0.465°C = -0.465°C
Therefore, the best choice is C, which rounds to -0.93°C because the provided answers are all in degrees Celsius and rounded to two decimal places.

A Solution Is Prepared by Dissolving 53