Question

$1.575$ g of oxalic acid ${\left(COOH\right)}_2\cdot x{H}_{2}O$ are dissolved in water and the volume is made upto $250$ mL. On titration, $16.68$ mL of this solution requires $25$ mL of $\frac{N}{15}$ $NaOH$ solution for complete neutralization. The value of $x$ is :

Answer 1

For the reaction between oxalic acid and $NaOH$,
Meq. of oxalic acid in $16.68$ mL $=$Meq. of $NaOH$ $=25\times \left(\frac{1}{15}\right)$
$\therefore$ Meq. of oxalic acid in $250$ mL$=25\times \frac{1}{15}\times \frac{250}{16.68}$
$\therefore \frac{1.575}{\frac{(90+18X)}{2}}\times 1000=24.98$
Hence, $x=2$.