At home, you dissolve 26.0 g of AlCl3 in 1.5 kg of water. What is the freezing point of this solution? (Normal f.p = 0.0 C, Kf= 1.86 C kg/mol)

mols AlCl3 = grams/molar mass. You know molar mass and grams, solve for mols.

Then m = molality = mols/kg solvent. You know kg solvent and mols, solve for m.

Then delta T = i*Kf*m
You know Kf and m; i for AlCl3 is 4. Solve for delta T.

Then subtract delta T from the normal freezing point to obtain the new freezing point.

To find the freezing point of the solution, we need to use the equation for calculating the freezing point depression:

ΔT = Kf * m

Where:
ΔT = change in freezing point (in degrees Celsius)
Kf = cryoscopic constant (in degrees Celsius per molality)
m = molality of the solution (in mol solute per kg of solvent)

First, we need to find the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.

Given:
Mass of AlCl3 = 26.0 g
Mass of water = 1.5 kg
Molar mass of AlCl3 = 133.34 g/mol (from periodic table)

Step 1: Calculate the number of moles of AlCl3
n(AlCl3) = mass / molar mass = 26.0 g / 133.34 g/mol ≈ 0.195 mol

Step 2: Calculate the molality (m)
m = n(AlCl3) / mass of water = 0.195 mol / 1.5 kg ≈ 0.13 mol/kg

Step 3: Substitute the values into the freezing point depression equation
ΔT = Kf * m = 1.86 °C kg/mol * 0.13 mol/kg ≈ 0.242 °C

Finally, to find the freezing point of the solution, we need to subtract the freezing point depression from the normal freezing point.

Freezing point = normal freezing point - ΔT
Freezing point = 0.0 °C - 0.242 °C ≈ -0.242 °C

Therefore, the freezing point of the AlCl3 solution is approximately -0.242 °C.