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

 $\begin{array}{l}\left[\begin{array}{r}\text{ Eq of }(\mathrm{COOH}{)}_2\cdot {\mathrm{nH}}_2\mathrm{O}\\ \text{ in }250\mathrm{mL}\text{ solution }\end{array}\right]=\left(\frac{1.575}{90+18\mathrm{n}}\right)\times 2\\ (\mathrm{n} \mathrm{factor} = 2) ........................... \left(\mathrm{i}\right)\\ \mathrm{mEq}\text{ of acid }=\mathrm{mEq}\text{ of }\mathrm{NaOH}\\ 16.68\mathrm{mL}\times \mathrm{N}=25\times \frac{1}{15}\\ \mathrm{N}\left(\text{ acid }\right)=0.099\approx 0.1{\mathrm{EqL}}^{-1}\\ =\frac{0.1\times 250}{1000} .......................\left(\mathrm{ii}\right)\\ =\frac{0.1}{4}\text{ Eq per }250\mathrm{mL}\end{array}$

Equating Eqs. (i) and (ii), we get

$\left(\frac{2\times 1.575}{90+18n}\right)=\frac{0.1}{4}$

Solve for n, n=2

Formula : (COOH)₂.2H₂O