What is the change in freezing point (delta T) of an aqueous solution that is 0.082 molality AlCl3?

ΔT = 0.61°C

Explanation:

That's the formula for the colligative property of freezing point depression:

ΔT = m . Kf . i

AlCl₃ is a ionic salt that be dissociated:

AlCl₃ → Al³⁺ + 3Cl⁻     i = 4

In solution, we can have 4 moles of ions, 1 mol of Al³⁺ plus 3 moles of chloride. (We assume 100 %ionization)

Let's replace data: ΔT = 0.082 m . 1.86 °C/m . 4

ΔT = 0.61°C

The change in freezing point of this aqueous solution is 0.61 °C

Explanation:

Step 1: Data given

Molality of the aqueous solution = 0.082 molal

freezing point of water = 0°C

Molal freezing point depression constant = 1.86 °C/m

Step 2: Calculate the freezing point

ΔT = i*Kf*m

⇒with ΔT = The freezing point depression = TO BE DETERMINED

⇒with i = The van't Hoff factor of AlCL3 = 4

⇒with Kf = the freezing point depression constant = 1.86 °C / molal

⇒with m = molality of the  aqueous solution = 0.082 molal

ΔT = 4 * 1.86 * 0.082

ΔT = 0.61 °C

Step 3: Calculate the freezing point of the aqueous solution

Freezing point solution = Freezing point of water - ΔT

Freezing point solution = 0° - 0.61 °C

Freezing point solution =  -0.61 °C

The change in freezing point of this aqueous solution is 0.61 °C

c) should be the answer.