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Question

The degree of dissociation (α) of a weak electrolyte, AxBy, is related to van’t Hof factor (i) by the expression

Options

A.$\alpha = \frac{i - 1}{x + y - 1}$
B.$\alpha = \frac{i - 1}{x + y + 1}$
C.$\alpha = \frac{x + y - 1}{i - 1}$
D.$\alpha = \frac{x + y + 1}{i - 1}$

Solution

$i = 1 + \alpha(n - 1) = 1 + \alpha(x + y - 1)$

$\therefore\alpha = \frac{i - 1}{x + y - 1}$

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