$50 m L$ of $0.5 M$ oxalic acid is needed to neutralize $25 m L$ of soidum hydroxide solution. The amount of $N a O H$ in $50 m L$ of the given sodium hydroxide solution is:

4 g

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20 g

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80 g

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10 g

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Solution

The correct option is A $4 g$
$H_{2} C_{2} O_{4} + 2 N a O H \to N a_{2} C_{2} O_{4} + 2 H_{2} O$

$m_{e q}$ of $H_{2} C_{2} O_{4} = m_{e q} o f N a O H$

$50 \times 0.5 \times 2 = 25 \times N_{N a O H} \times 1$

$∴ M_{N a O H} = N_{N a O H} = 2 M$ ; since n-factor is 1

Now, $1000 m l$ solution $= 2 \times 40 g r a m N a O H$

$∴ 50 m l$ solution $= 4 g r a m N a O H$ .

Hence, the correct option is A.

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