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.

For ideal solution, we have

$$\Delta $$H_mix = 0, $$\Delta $$V_mix = 0

Now U_mix = ∆H_mix – P$$\Delta $$V_mix

$$ \therefore $$ $$\Delta $$U_mix = 0

Also, for an ideal solution,

p_A = x_Ap_A^o, p_B = x_Bp_B^o

$$ \therefore $$ $$\Delta $$p = p_observed – p_calculated = 0

$$\Delta $$G_mix = $$\Delta $$H_mix – T$$\Delta $$S_mix

For an ideal solution, $$\Delta $$S_mix $$ \ne $$ 0

$$ \therefore $$ $$\Delta $$G_mix $$ \ne $$ 0

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.