The mole fraction of cation (Al)=0.4
Further explanation
Given
1 M of Aluminium sulphate
Required
The mole fraction of cation
Solution
Ionization of the Aluminum sulfate solution( assume 1 L solution ) :
mol Al₂(SO₄)₃ = M x V = 1 M x 1 L = 1 mol
Al₂(SO₄)₃⇒2Al³⁺ + 3SO₄²⁻
1 mol 2 mol 3 mol
From this equation, total mol in solution = 2+3 = 5 moles
Mol fraction Al(as a cation) :
= 2/5=0.4
