`3.150 g` of oxalic acid `[(COOH)_(2).xH_(2)O]` are dissolved in water and volume made up to `500 mL`. On titration `28 mL` of this solution required `35 mL` of `0.08N NaOH` solution for complete neutralization. Find the value of `x`.

Answer

2

Explanation

Meq.of oxalic acid in `28 mL= "Meq.of" NaOH`
`=0.08xx35`
`therefore` Meq.of oxalic acid in `500 mL`
`=(500)/(28)xx35xx0.08=50`
`therefore (3.150)/(m//2)xx1000=50 rArr m =126`
`therefore H_(2)C_(2)O_(4).xH_(2)O=90+18x=126`
or `x=2`