How to Calculate Freezing Point Depression with Al(NO3)3?

What is the freezing point (°C) of a solution prepared by dissolving 15.6 g of Al(NO3)3 in 150 g of water?

Can you explain how to calculate the freezing point depression of the solution?

Answer:

The freezing point of the solution prepared by dissolving 15.6 g of Al(NO3)3 in 150 g of water is approximately -0.905 °C.

To find the freezing point of the solution, we need to calculate the molality (m) of the solution, which is defined as the moles of solute per kilogram of solvent.

First, we need to convert the mass of Al(NO3)3 to moles. The molar mass of Al(NO3)3 is 213.0 g/mol.

moles of Al(NO3)3 = mass / molar mass

= 15.6 g / 213.0 g/mol

= 0.073 moles

Next, we calculate the molality (m) using the moles of solute and the mass of the solvent.

molality (m) = moles of solute / mass of solvent (in kg)

= 0.073 moles / 0.150 kg

= 0.487 molal

Given that the cryoscopic constant (Kf) for water is 1.86 °C/m, we can use the freezing point depression equation to find the freezing point depression (ΔTf).

ΔTf = Kf * m

= 1.86 °C/m * 0.487 molal

= 0.905 °C

The freezing point depression represents the difference between the freezing point of the pure solvent (water) and the freezing point of the solution. Therefore, to find the freezing point of the solution, we subtract the freezing point depression from the freezing point of pure water (0 °C).

Freezing point = 0 °C - 0.905 °C

= -0.905 °C

Therefore, the freezing point of the solution prepared by dissolving 15.6 g of Al(NO3)3 in 150 g of water is approximately -0.905 °C.