103k views
0 votes
A certain liquid X has a normal freezing point of 3.30 °C and a freezing point depression constant K = -3.27 °c-kg-mol⁻¹ . A solution is prepared by dissolving some ammonium chloride (NH₄CI) in 400. g of X. This solution freezes at 1.6 °C. Calculate the mass of NH₄Cl that was dissolved.

User Philar
by
8.0k points

1 Answer

2 votes

Final answer:

To calculate the mass of NH₄Cl dissolved, the freezing point depression formula is used. After finding the change in freezing point, the molality is calculated, and then the moles of NH₄Cl are determined. Multiplying this by the molar mass of NH₄Cl yields the mass dissolved, approximately 11.121 g.

Step-by-step explanation:

The student has asked to calculate the mass of ammonium chloride (NH₄Cl) dissolved in a liquid with a given normal freezing point and a freezing point depression constant. To find this, we use the freezing point depression formula ΔTf = Kf x m, where ΔTf is the change in freezing point, Kf is the freezing point depression constant, and m is the molality of the solution.

First, we calculate the change in freezing point (ΔTf) of the solvent, which is the difference between the normal freezing point of liquid X and the observed freezing point:

ΔTf = 3.30 °C - 1.6 °C = 1.7 °C.

Next, we find the molality (m) of the solution:

m = ΔTf / Kf = 1.7 °C / (-3.27 °C kg mol⁻¹) = -0.51957 mol/kg.

Note that the negative sign in the freezing point depression constant and ΔTf cancel each other out, giving a positive molality. Since we have 400 g of liquid X, which is 0.4 kg, we can calculate the number of moles of NH₄Cl:

number of moles = molality x mass of solvent = 0.51957 mol/kg x 0.4 kg = 0.20783 mol.

Finally, to find the mass of NH₄Cl dissolved, we use the molar mass of NH₄Cl (53.49 g/mol):

mass of NH₄Cl = number of moles x molar mass = 0.20783 mol x 53.49 g/mol = 11.121 g.

Therefore, approximately 11.121 g of NH₄Cl was dissolved in the solvent.

User Jay Otterbein
by
8.4k points