Which one of the following electrolytes has the same value of van't Hoff's factor (i) as that of Al₂(SO₄)₃ (if all are 100% ionized)?

Van't Hoff factor i = (number of solute particles present in solution) / (theoretical number of solute particles due to solution of non-electrolyte) = n(observed) / n(theoretical).

1 molecule of Al₂(SO₄)₃ ionizes in solution to produce 5 ions: 2Al³⁺ + 3SO₄²⁻. Therefore, the van't Hoff factor (i) for Al₂(SO₄)₃ is 5.

Let's examine the options:

• K₂SO₄ ionizes into 3 ions: 2K⁺ + SO₄²⁻. Therefore, i = 3.
• Al(NO₃)₃ ionizes into 4 ions: Al³⁺ + 3NO₃⁻. Therefore, i = 4.
• K₃[Fe(CN)₆] ionizes into 4 ions: 3K⁺ + [Fe(CN)₆]³⁻. Therefore, i = 4.
• K₄[Fe(CN)₆] ionizes into 5 ions: 4K⁺ + [Fe(CN)₆]⁴⁻. Therefore, i = 5.

Therefore, K₄[Fe(CN)₆] has the same van't Hoff factor (i = 5) as Al₂(SO₄)₃.