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Standard XII

Chemistry

Mole Concept

1.575 g of ox...

Question

$1.575 g$ of oxalic acid $(C O O H)_{2} \cdot x H_{2} O$ are dissolved in water and the volume made up to $250 m L$ . On titration $16.68 m L$ of this solution requires $26 m L$ of $\frac{N}{15} N a O H$ solution for complete neutralisation. Calculate $x$ .

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Solution

$25 m L$ of $\frac{N}{15} N a O H$ solution $\equiv 25 m L$ of $\frac{N}{15}$ oxalic acid solution
Mass of oxalic acid present in $25 m L$ of $\frac{N}{15}$ oxalic acid solution
$= \frac{N \times E \times V}{1000} = \frac{1 \times (90 + 18 x) \times 25}{15 \times 2 \times 1000}$
$= \frac{(90 + 18 x)}{1200} g$
Actually $\frac{(90 + 18 x)}{1200} g$ oxalic acid is present in $16.68 m L$ solution.
$250 m L$ of the solution contains oxalic acid
$= \frac{(90 + 18 x) \times 250}{1200 \times 16.68} = 1.575 (g i v e n)$
or $90 + 18 x = \frac{1.575 \times 1200 \times 16.68}{250} = 126$
or $18 x = 126 - 90 = 36$
$x = 2$ .

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1.575 g of oxalic acid (COOH)2⋅xH2O are dissolved in water and the volume made up to 250 mL. On titration 16.68 mL of this solution requires 26 mL of N15NaOH solution for complete neutralisation. Calculate x.

$1.575 g$ of oxalic acid $(C O O H)_{2} \cdot x H_{2} O$ are dissolved in water and the volume made up to $250 m L$ . On titration $16.68 m L$ of this solution requires $26 m L$ of $\frac{N}{15} N a O H$ solution for complete neutralisation. Calculate $x$ .