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Types of Redox Reactions

1.575 g of ox...

Question

$1.575 g$ of oxalic acid $(H_{2} C_{2} O_{4} . x H_{2} O)$ are dissolved in water and the volume made upto $250 m l$ . On titration, $16.68 m l$ of this solution requires $25 m l$ of $\frac{N}{15}$ $N a O H$ solution for complete neutralization. Calculate the value $x$ .

A

3

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B

2

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C

4

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D

5

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Solution

The correct option is B $2$
Milliequivalents of oxalic acid in $16.68$ ml $=$ Milliequivalents of $N a O H$ $= 25 \times \frac{1}{15}$

Milliequivalents of oxalic acid in $250$ ml $=$ $25 \times \frac{1}{15} \times \frac{250}{16.68} = 24.98$

Therefore, $\frac{1.575}{\frac{(90 + 18 x)}{2}} \times 1000 = 24.98$
$∴ x = 2$

Therefore, the value of $x$ is $2$ .

Hence, option B is correct.

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