5.33 gram of $CrCl_3 \cdot 6H_2O$ (1:3 electrolyte) is passed through cation exchanger. The resulting solution is then treated with an excess of $AgNO_3$, leading to formation of 8.61 gran of precipitate. Calculate : $\frac{\text{number of moles of complex reacted}}{\text{number of moles of AgCl precipitated}} \times 100$
Quantitative analysis of coordination compound dissociation using ion exchange and precipitation.
CALCULATION: 1. Molar mass of $CrCl_3 \cdot 6H_2O$ is $266.5 \text{ g/mol}$. 2. Moles of complex used = $5.33 / 266.5 = 0.02 \text{ mol}$. 3. The complex is a 1:3 electrolyte, meaning it is $[Cr(H_2O)_6]Cl_3$. All 3 Chloride ions are ionizable. 4. Passing through a cation exchanger replaces $[Cr(H_2O)_6]^{3+}$ with $3H^+$, but the $3Cl^-$ remain in solution as $3HCl$. 5. Reacting with $AgNO_3$ precipitates all chloride as $AgCl$. 6. Moles of $AgCl$ = $8.61 / 143.5 = 0.06 \text{ mol}$. 7. Required value = $(0.02 / 0.06) \times 100 = (1/3) \times 100 = 33.33$. The integer answer is 33.
