A dilute solution of K ₂ HgI ₄ reagent is 95 % ionised. What would be the approximate value of its van't Hoff factor?

Step-by-step solution

The dissociation of K ₂ HgI ₄ in a dilute solution is represented as: K ₂ HgI ₄ 2 K ⁺ + HgI ₄²⁻ The number of ions produced per formula unit is n = 3 . The van't Hoff factor i is given by the formula i = 1 + (n - 1) , where is the degree of dissociation. Given = 0.95 and n = 3 , we substitute these values into the formula: i = 1 + 0.95(3 - 1) i = 1 + 0.95(2) i = 1 + 1.90 i = 2.90 Answer: 2.90