A 0.050 M solution of AlCl3 had an observed osmotic pressure of 3.85 atm at 20 ∘C . Calculate the van't Hoff factor i for AlCl3.

Van't Hoff factor, for this solution is 3.2

Explanation:

Osmotic pressure formula → π = M . R . T . i

π = Pressure → 3.85 atm

M = Molarity → 0.05mol/L

R = Constant for the Ideal Gases → 0.082 L.atm / mol.K

T = Absolute T° (K) → T°C + 273

20°C + 273 = 293K

Let's replace data:

3.85 atm = 0.05 mol/L . 0.082 L.atm/mol.K . 293K . i

3.85 atm / 0.05  . 0.082 atm . 293 = i

3.20 = i

If we assume complete ionization, i must be ≅ 4

AlCl₃ →  Al³⁺  +  3Cl⁻

We have 1 mol of aluminum cation and 3 moles of chloride, 4 total moles.