$0.1 M$ aqueous solution of $M g C l_{2}$ is at 300K and 4.92 atm. What will be the percentage of ionization of the salt?

59 %

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69 %

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79 %

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49 %

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Solution

The correct option is D $49 %$
$π = i c R t$

$α = \frac{i - 1}{n - 1}$

$π = i c R t$

$4.92 = i \times 0.1 \times 0.0821 \times 300$

$i = 1.99$

$α = \frac{i - 1}{n - 2} = \frac{1.99 - 1}{3 - 1} = \frac{99}{2} = 49$ (approxiamately)

Therefore, $%$ of ionization = $49 %$

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