A $0.15 M$ aqueous solution of $K C l$ exerts an osmotic pressure of $6.8$ atm at $310 K$ . Calculate the degree of dissociation of $K C l$ . $(R = 0.0821 L i t . a t m K^{- 1} {m o l}^{- 1})$ .

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Solution

The osmotic pressure $π = i C R T$
$6.8 a t m = i \times 0.15 0.0821 L i t a t m / (K m o l) \times 310 K$
$i = 1.78$
One molecule of KCl dissociates to give two ions
$K C l \to K^{+} + C l^{-}$
$n = 2$
The degree of dissociation of KCl $α = \frac{i - 1}{n - 1}$
$α = \frac{1.78 - 1}{2 - 1}$
$α = 0.78$

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