JEE Chemistry 2023 Question with Solution

Given:

• Molarity of $\mathrm{[Cr(H_2O)_6]Cl_2}$ solution = $0.01 \, \mathrm{M}$
• Volume of solution = $20 \, \mathrm{mL} = 0.020 \, \mathrm{L}$
• Molarity of $\mathrm{AgNO_3}$ = $0.1 \, \mathrm{M}$

Find: The volume of $\mathrm{AgNO_3}$ needed for complete precipitation of chloride ions as $\mathrm{AgCl}$.

In $\mathrm{[Cr(H_2O)_6]Cl_2}$, both chloride ions are outside the coordination sphere. Therefore, each formula unit gives $2$ chloride ions.

Moles of $\mathrm{[Cr(H_2O)_6]Cl_2}$:

$$\text{moles} = M \times V = 0.01 \times 0.020 = 2.0 \times 10^{-4} \; \mathrm{mol}$$

Moles of chloride ions:

$$\text{moles of } \mathrm{Cl^-} = 2 \times 2.0 \times 10^{-4} = 4.0 \times 10^{-4} \; \mathrm{mol}$$

Precipitation reaction:

$$\mathrm{Ag^+ + Cl^- \rightarrow AgCl(s)}$$

This is a $1:1$ mole ratio, so moles of $\mathrm{AgNO_3}$ required are also $4.0 \times 10^{-4} \, \mathrm{mol}$.

Now,

$$V = \frac{\text{moles}}{\text{molarity}} = \frac{4.0 \times 10^{-4}}{0.1} = 4.0 \times 10^{-3} \; \mathrm{L}$$

Converting to mL:

$$4.0 \times 10^{-3} \; \mathrm{L} = 4 \; \mathrm{mL}$$

Therefore, the required volume is $4 \, \mathrm{mL}$.