Question

the freezing point of 0.2 molal K2SO4 is -1.1°c calculate percentage degree of dissociation of K2SO4(kf for water is 1.86°)​

Answer 1

percentage of degree of dissociation is 97.8 %

It has given that the freezing point of 0.2 molal K₂SO₄ is -1.1 °C.

we have to calculate percentage degree of dissociation of K₂SO₄.

let α is degree of dissociation of Potassium sulphate.

K₂SO₄ ⇔2K⁺ + SO₄¯ ¯

1 - α 2α α

now i = 1 - α + 2α + α = 1 + 2α

using formula, ∆T_f = i × k_f × m

here m is molality of potassium sulphate. K_f is constant i.e., K_f for water is 1.86 °C/molal

⇒1.1 = (1 + 2α) × 1.86 × 0.2

⇒5.5/1.86 = 1 + 2α

⇒1 + 2α = 2.956

⇒α = 0.978

therefore percentage of degree of dissociation is 97.8 %