A solution containing 0.319 gm of complex ${\mathrm{CrCl}}_3 . 6{\mathrm{H}}_2\mathrm{O}$ was passed through cation exchanger and the solution given out was neutralized by 28.5 ml of 0.125 $\mathrm{M}$ $\mathrm{NaOH}$. The correct formula of the complex will be : [Molecular weight of complex = 266.5)

The resulting product is $\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_3$, with three chloride ions outside the complex sphere for each mole.
Given, Complex's molecular weight is = 266.5.
Complex mass is 0.319 grams.

${ }^{\mathit{n}}\mathit{(}CrC{l}_3\cdot 6H_{2}O)=\frac{0.319}{266.5}$

$n_{(}CrC{l}_3\cdot 6H_{2}O)=1.197m\mathit{ }mol$

The moles of acid are,

$\Rightarrow 1.197\times 3=3.5625$