1.575 g of oxalic acid (COOH)_(2).xH_(2)O are dissolved in water and the volume made up to 250 mL. On titration 16.68 mL of this solution requires 25 mL of N/15 NaOH solution for complete neutralisation. Calculate the value of x.

1.575 g of oxalic acid (COOH)_(2).xH_(2)O are dissolved in water and t

1.	1.575 g of oxalic acid (COOH)_(2).xH_(2)O are dissolved in water and the volume made up to 250 mL. On titration 16.68 mL of this solution requires 25 mL of N/15 NaOH solution for complete neutralisation. Calculate the value of x.
Answer» Solution :Molecular mass of oxalic ACID, `(COOH)_2 xH_2O = 24 + 62 + 2 + 18x =90 + 18x` Since oxalic acid is a dibasic acid, Eq. mass of oxalic acid `=("Molecular mass")/2` `=(90 + 8x)/2 = 45 + 9x` Normality of oxalic acid solution can be calculated from the following relation : `w=(NEV)/1000` SUBSTITUTING the values, we get `1.575 =(N xx (45 + 9x) xx 250)/1000` or `N =(1.575 xx 1000)/((45 + 9x) xx 250) = (6.3)/(45 + 9x)` According to the normality relation, `underset("oxalic acid")(N_(1)V_(1)) = underset("NaOH")(N_(2)V_(2))` `6.3/(45 + 9x) xx 16.68 = 1/15 xx 25` `45 + 9x = (6.3 xx 16.68 xx 15)/25` `=63.05 = 63` `9x = 63 - 45 = 18` or `x=18/9 =2`