1.575 g, of hydrated oxalic acid (COOH)2.nH2O is dissolved in water and the solution is made to 25OmL. On titration, 16.68 mL of this solution is required for neutralisation of 25 mL of N/15 NaOH. The value of water crystallisation, i.e. , n, is

Mw of (COOH)2. nH2O = 90 + 18n  Eq of (COOH)2⋅nH2O in 250mL solution =1.57590+18n×2                                                 (n factor = 2) ........................... (i)mEq of acid =mEq of NaOH16.68mL×N=25×115N( acid )=0.099≈0.1EqL−1=0.1×2501000                                               .......................(ii)=0.14 Eq per 250mLEquating Eqs. (i) and (ii), we get2×1.57590+18n=0.14Solve for n, n=2Formula : (COOH)2.2H2O