What volume of $0.1 M H_{2} S O_{4}$ is required in $l i t r e s$ to neutralize completely $1 l i t r e$ of a solution containing $20 g$ of $N a O H$ ?

5.0

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0.5

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2.5

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10.0

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Solution

The correct option is B $2.5$
$M o l a r i t y o f N a O H = \frac{W e i g h t \times 1000}{m o l a r m a s s \times V} = \frac{20 \times 1000}{40 \times 1000} = 0.5$

Reaction: $H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O$

If 1 is for $H_{2} S O_{4}$ and 2 is for $N a O H$

$\frac{M_{1} V_{1}}{1} = \frac{M_{2} V_{2}}{2}$

$0.1 \times V_{1} = \frac{0.5 \times 1}{2}$

$∴ V_{1} = 2.5 L$ .

Hence, option $C$ is correct.

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