5.33 g of CrCl$$_3$$·6H$$_2$$O, which is a 1 : 3 electrolyte, is dissolved in water and is passed through a cation exchanger. The chloride ions in the eluted solution, on treatment with AgNO$$_3$$ results in 8.61 g of AgCl. The ratio of moles of complex reacted and moles of AgCl formed is __________ × 10$$^{-2}$$. (Nearest integer) [Molar mass in g mol$$^{-1}$$ Cr : 52, Ag : 108, Cl : 35.5, H : 1, O : 16]

Molar mass of the given compound $$CrCl_3\cdot 6H_2O$$:

$$M = 52 + 3(35.5) + 6(18) = 52 + 106.5 + 108 = 266.5\ \text{g mol}^{-1}$$

Moles of the complex taken:

$$n_{\text{complex}} = \frac{5.33}{266.5} = 0.0200\ \text{mol}$$

The compound behaves as a $$1:3$$ electrolyte, so all three chloride ions remain outside the coordination sphere. A cation-exchange resin removes the cationic complex, leaving these three $$Cl^-$$ ions in the eluted solution.

The eluate is treated with $$AgNO_3$$, giving $$AgCl$$.

Molar mass of $$AgCl$$: $$108 + 35.5 = 143.5\ \text{g mol}^{-1}$$

Moles of $$AgCl$$ obtained (equal to moles of $$Cl^-$$):

$$n_{AgCl} = \frac{8.61}{143.5} = 0.0600\ \text{mol}$$

Ratio of moles of complex reacted to moles of $$AgCl$$ formed:

$$\frac{n_{\text{complex}}}{n_{AgCl}} = \frac{0.0200}{0.0600} = \frac{1}{3} = 0.333$$

Expressing this as $$\times 10^{-2}$$:

$$0.333 = 33.3 \times 10^{-2}$$

Nearest integer = $$33$$.

Answer: 33