Ba(OH)₂ is a strong electrolyte and undergoes cent percent dissociation in a dilute aqueous solution.

Ba(OH)_2(aq) $$ \to $$ Ba²⁺_(aq) + 2OH^–_(aq)

Thus, van’t Hoff factor i = 3.

Ai₂(SO₄)₃ ⇌ 2Al⁺³ + 3SO₄^2–

van't Hoff factor, i = 5

K₂SO₄ ⇌ 2K⁺ + SO₄^2–

van't Hoff factor, i = 3

K₃[Fe(CN)₆] ⇌ 3K⁺ + [Fe(CN)₆]^2-

van't Hoff factor, i = 4

Al(NO₃)₃ ⇌ Al³⁺ + 3NO₃^–

van't Hoff factor, i = 4

K₄[Fe(CN)₆] ⇌ 4K⁺ + [Fe(CN)₆]^4–

van't Hoff factor, i = 5

For an ideal solution, $$\Delta $$S_mix > 0 while $$\Delta $$H_mix, $$\Delta $$V_mix and $$\Delta $$P all are 0.

Number of moles of water in 1000 g = $${{1000} \over {18}}$$ = 55.56 mol

Thus, mole fraction of solute = $${1 \over {1 + 55.56}}$$ = 0.0177