Question

The freezing point of water is 0.00°C at 1 atmosphere.
If 13.40 grams of ammonium sulfide, (68.20 g/mol), are dissolved in 153.2 grams of water. The molality of the solution is______ m.
The freezing point of the solution is ______ °C.

Answer 1

Answer:

Molality is: 1,283m

Freezing point of the solution is: -7,159°C.

Explanation:

Molality is defined as moles of solute per kg of solvent

Moles of ammonium sulfide (solute):

13,40g × (1mol / 68,20g) = 0,1965moles

Molality is:

0,1965moles / 0,1532kg Water = 1,283m

The freezing point depression formula is:

ΔT = Kf × m × i

Where ΔT is change in temperature, k is solvent constant (1.86 °C/m for water), m is molality (1,283m) and i is Van't Hoff factor (3 for ammonium sulfide -van't hoff factor is the number of ions which is dissociates in the solution, for (NH₄)₂SO₄ ions are 3-. Replacing:

ΔT = 1,86°C/m × 1,283m × 3

ΔT = 7,159°C

As freezing point of water is 0°C, freezing point of the solution is 0°C - 7,159°C = -7,159°C

I hope it helps!