1.Relationship between van't Hoff's factor (iii) and degree of dissociation (α\alphaα) isa.i=α−1n′−1i = \frac{\alpha - 1}{n' - 1}i=n′−1α−1b.i=α−11−n′i = \frac{\alpha - 1}{1 - n'}i=1−n′α−1c.α=1−in′−1\alpha = \frac{1 - i}{n' - 1}α=n′−11−id.α=i−1n′−1\alpha = \frac{i - 1}{n' - 1}α=n′−1i−1Login to continueOnly logged in users canattempt or see the solution.
