The normality of $4 %$ (wt/vol.) $N a O H$ is:

0.1

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1.0

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0.05

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0.01

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Solution

The correct option is D $0.01$
$4$ % (wt/vol) means that $4 g$ of solute is dissolved in $100 m l$ of solvent

$4 g o f s o l u t e ⟶ 100 m l o f s o l v e n t$

$N o r m a l i t y (N) = \frac{N o . o f e q u i v a l e n t s o f s o l u t e}{V o l u m e o f s o l u t i o n i n l i t r e}$

$N o . o f e q u i v a l e n t s = \frac{4}{40} = \frac{1}{10}$

$N o r m a l i t y (N) = \frac{1 / 10}{100 / 1000} = 1 N$

The normality of $4$ % (wt/vol) NaOH is $1.0 N$

Hence, the correct option is $B .$

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