Practicing Success
The van't Hoff factor for a 0.1 M Al2(SO4)3 solution is 4.2. What is the degree of dissociation? |
80% 90% 78% 83% |
80% |
Al2(SO4)3 ⇔ 2Al2+ + 3SO43- 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% |