The van't Hoff factor for a 0.1 M Al2(SO4)3 solution is 4.2. What is…

Al₂(SO₄)₃ ⇔ 2Al²⁺ + 3SO₄^3-

   1                 0          0        (moles before dissociation)

  1-α               2α        3α      (moles after dissociation)

Total moles after dissociation = 1-α + 2α + 3α = 1+4α

i = \(\frac{\text{Total moles after dissociation}}{\text{Initial no of moles}}\)

4.2 = \(\frac{1+4α}{1}\)

4.2 = 1+4α

3.2 = 4α

α = \(\frac{3.2}{4}\) = 0.8

Degree of dissociation = α x 100 = 0.8 x 100 = 80%