The n-factor of $H_{2} S O_{4}$ in the following reaction is:
$H_{2} S O_{4} + K O H \to K H S O_{4} + H_{2} O$

1

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2

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\frac{1}{2}

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3

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Solution

The correct option is A $1$
$N -$ factor is the basicity of acid, i.e. the number of hydrogen atoms replaceable by a base in a particular acid.
In given reaction:-
$H_{2} S O_{4} + K O H ⟶ K H S O_{4} + H_{2} O$
$N -$ factor of $H_{2} S O_{4}$ is $1$ , as only one Hydrogen of acid is replaced by $K O H$ .

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