The degree of association $(α)$ is given by the expression

$α = \frac{n (i - 1)}{1 - n}$

$α = \frac{i (n - 1)}{1 + n}$

$α = \frac{i (n + 1)}{1 - n}$

$α = \frac{i (n + 1)}{n - 1}$

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The correct Answer is:A

$n A 1 mole \Leftrightarrow A_{n}$
$1 - α \frac{α}{n}, Total = 1 - α + \frac{α}{n}$
$∴ i = 1 - α + \frac{α}{n} or 1 - i = α - \frac{α}{n} = α (1 - \frac{1}{n})$
$= α (\frac{n - 1}{n})$
$or α = \frac{n}{(n - 1)} (i - i) = \frac{n (i - 1)}{(1 - n)}$

Updated on:21/07/2023

Class 12 CHEMISTRY SOLUTIONS

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The degree of association $(α)$ is given by the expression

The correct Answer is:A

$n A 1 mole \Leftrightarrow A_{n}$
$1 - α \frac{α}{n}, Total = 1 - α + \frac{α}{n}$
$∴ i = 1 - α + \frac{α}{n} or 1 - i = α - \frac{α}{n} = α (1 - \frac{1}{n})$
$= α (\frac{n - 1}{n})$
$or α = \frac{n}{(n - 1)} (i - i) = \frac{n (i - 1)}{(1 - n)}$

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nA1 mole⇔An 1−α αn, Total=1−α+αn ∴i=1−α+αnor1−i=α−αn=α(1−1n) =α(n−1n) orα=n(n−1)(i−i)=n(i−1)(1−n)

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