Van't Hoff factor of aqeous $K_{2} S O_{4}$ at infinite dilution has value equal to:

1

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2

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3

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Between

2

and

3

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Solution

The correct option is C $3$
At infinite dilution, $K_{2} S O_{4}$ will ionise $100 % .$
$K_{2} S O_{4} \to 2 K^{+} + S O_{4}^{2 -}$
Therefore, at infinite dilution, van't Hoff factor, $i = 3.$

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