$250 m l$ aqueous solution of $6.3 g$ of oxalic acid dihydrate is made. The volume of $0.1 N N a O H$ required to completely neutralise $10 m l$ of this solution is:

20 m l

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40 m l

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10 m l

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4 m l

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Solution

The correct option is B $40 m l$
The molecular mass of oxalic acid dihydrate ( $C O O H - C O O H . 2 H_{2} O$ )=126

Gram equivalent= $\frac{126}{2} = 63$ ; (basicity=2)

Normality of oxalic acid $= \frac{m a s s \times 1000}{g r a m e q u i v a l e n t \times v o l u m e}$ $= \frac{6.3 \times 1000}{63 \times 250} = 0.4 N$

$N_{1} V_{1} = N_{2} V_{2}$
Where, 1 represents for $H_{2} S O_{4}$ and 2 reresents for $N a O H$ .

$∴ 10 \times 0.4 = V \times 0.1$ .

$∴ V = \frac{10 \times 0.4}{0.1} = 40 m l$

Hence, option $B$ is correct.

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Similar questions

Q. An aqueous solution of 6.3 g of oxalic acid dihydrate

(C_{2} H_{2} O_{4} .2 H_{2} O)

has a volume equal to 250 mL. The volume of 0.1 M

N a O H

required to completely neutralize 10 mL of this solution is:

A known mass of oxalic acid dihydrate is dissolved in water to make 100 ml solution. 10 ml of this solution is diluted to 100 ml. 20 mL of the solution requires 20 mL of 0.1 N NaOH for complete neutralisation. (eq. mass of oxalic acid dihydrate = 63). The mass of oxalic acid in gram in the original solution is

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250 ml aqueous solution of 6.3 g of oxalic acid dihydrate is made. The volume of 0.1 N NaOH required to completely neutralise 10 ml of this solution is:

$250 m l$ aqueous solution of $6.3 g$ of oxalic acid dihydrate is made. The volume of $0.1 N N a O H$ required to completely neutralise $10 m l$ of this solution is: