1.In a solution, if three molecules of solute undergo association as: 3A⇌A33A \rightleftharpoons A_33A⇌A3, then to evaluate degree of association 'α\alphaα' the formula will be:-a.α=3(1−i)2\alpha = \dfrac{3(1 - i)}{2}α=23(1−i)b.α=i−12\alpha = \dfrac{i - 1}{2}α=2i−1c.α=3(i2−1)2\alpha = \dfrac{3(i^2 - 1)}{2}α=23(i2−1)d.α=i−13\alpha = \dfrac{i - 1}{3}α=3i−1Login to continueOnly logged in users canattempt or see the solution.
