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Percentage Composition

The volume of...

Question

The volume of 0.1 M NaOH required for the neutralization of 20ml of 0.05 M $H_{2} C_{2} O_{4}$

20ml

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30ml

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40ml

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50ml

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Solution

The correct option is A 20ml
$2 N a O H + H_{2} C_{2} O_{4} \to N a_{2} C_{2} O_{4} + 2 H_{2} O \frac{M_{1} V_{1}}{n_{1}} = \frac{M_{2} V_{2}}{n_{2}} V_{2} = \frac{0.05 \times 20}{1} \times \frac{2}{0.1} = 20 m l$

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