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Abnormal Colligative Properties

10 mL of Na...

Question

$10$ mL of $N a H C_{2} O_{4}$ is oxidized by $10$ mL of $0.02 M$ $M n O_{4}^{⊝}$ . Hence, $10$ mL of $N a H C_{2} O_{4}$ is neutralized by:

A

10

mL of

0.1 M N a O H

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B

10

mL of

0.02 M N a O H

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C

10

mL of

0.1 M C a (O H)_{2}

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D

10

mL of

0.05 N B a (O H)_{2}

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Solution

The correct option is A $10$ mL of $0.1 M N a O H$
In $N a H C_{2} O_{4}, C_{2} O_{4}^{2 -}$ is oxidised to $C O_{2}$ and $H^{⨁}$ is neutralized.
Equivalent weight of $N a H C_{2} O_{4}$ $= \frac{M}{2}$
$10 \times N_{1} (N a H C_{2} O_{4}) = 10 \times 0.1 N$ $(M n O_{4}^{⊝})$
$N_{1} = 0.1 N$
Hence, $10$ mL of $N a H C_{2} O_{4}$ is neutralised by $10$ mL of $0.1 M N a O H$ .

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