If A2B is 30% ionised in an aqueous solution, then the value of van’t Hoff factor (i) is _____ ×10-1. [2025]
(16)
A2B⇌2A++B2-
i=1+(n-1)α
One molecule of A2B dissociates into three particles, so n = 3.
α = 0.3 as per question.
i=1+(3-1)×0.3
i=1.6=16×10-1