If a compound X ₃ Y is 60 % ionised in aqueous medium, what is its van't Hoff factor value?

Step-by-step solution

The compound X₃Y dissociates in aqueous medium as follows: X₃Y 3X⁺ + Y³⁻ The number of ions produced per formula unit is n = 3 + 1 = 4 . The degree of dissociation is given as = 60 % = 0.6 . The van't Hoff factor i is calculated using the formula i = 1 + (n - 1) . Substituting the values n = 4 and = 0.6 into the formula: i = 1 + 0.6(4 - 1) i = 1 + 0.6(3) i = 1 + 1.8 i = 2.8 Answer: 2.8