The ratio of the numer of moles of `AgNO_(3), Pb(NO_(3))_(2) and Fe(NO_(3))_(3)` required for coagulation of a difinite amount of a colloidal sol of silver iodide prepared by maxing `Ag NO_(3)` with excess of KI will be

Options

A. `1:2:3`
B. `3:2:1`
C. `6:3:2`
D. `2:3:6`

Explanation

To solve the problem, we need to determine the ratio of the number of moles of \( \text{AgNO}_3 \), \( \text{Pb(NO}_3)_2 \), and \( \text{Fe(NO}_3)_3 \) required for the coagulation of a colloidal sol of silver iodide (\( \text{AgI} \)).

### Step-by-Step Solution:

1. **Understanding the Formation of Colloidal Silver Iodide**:
- When \( \text{AgNO}_3 \) is mixed with excess \( \text{KI} \), a precipitate of silver iodide (\( \text{AgI} \)) is formed:
\[ \text{AgNO}_3 + \text{KI} \rightarrow \text{AgI} \downarrow + \text{KNO}_3 \]
- The \( \text{AgI} \) formed is a colloidal sol.

2. **Coagulation of Colloidal Sol**:
- Coagulation of colloidal particles occurs when oppositely charged ions are introduced to neutralize the charge on the colloidal particles.
- The ions from \( \text{AgNO}_3 \), \( \text{Pb(NO}_3)_2 \), and \( \text{Fe(NO}_3)_3 \) can be used for this purpose.

3. **Determining the Charge of the Colloidal Particles**:
- The \( \text{AgI} \) colloidal particles are negatively charged due to the presence of iodide ions (\( \text{I}^- \)).
- To coagulate these negatively charged particles, we need cations.

4. **Cations from the Salts**:
- \( \text{AgNO}_3 \) provides \( \text{Ag}^+ \) ions.
- \( \text{Pb(NO}_3)_2 \) provides \( \text{Pb}^{2+} \) ions.
- \( \text{Fe(NO}_3)_3 \) provides \( \text{Fe}^{3+} \) ions.

5. **Calculating the Ratio**:
- The charge neutrality required for coagulation can be considered:
- 1 mole of \( \text{Ag}^+ \) neutralizes 1 mole of \( \text{I}^- \).
- 1 mole of \( \text{Pb}^{2+} \) neutralizes 2 moles of \( \text{I}^- \).
- 1 mole of \( \text{Fe}^{3+} \) neutralizes 3 moles of \( \text{I}^- \).
- Therefore, the moles required for coagulation can be expressed as:
- For \( \text{AgNO}_3 \): 1 mole of \( \text{Ag}^+ \) for 1 mole of \( \text{I}^- \)
- For \( \text{Pb(NO}_3)_2 \): 0.5 moles of \( \text{Pb(NO}_3)_2 \) for 1 mole of \( \text{I}^- \)
- For \( \text{Fe(NO}_3)_3 \): \( \frac{1}{3} \) moles of \( \text{Fe(NO}_3)_3 \) for 1 mole of \( \text{I}^- \)

6. **Finding the Common Ratio**:
- To find the ratio of the moles of \( \text{AgNO}_3 : \text{Pb(NO}_3)_2 : \text{Fe(NO}_3)_3 \), we can express them in terms of a common base:
- Let’s take the common base as 6:
- \( \text{AgNO}_3 \): 6 moles
- \( \text{Pb(NO}_3)_2 \): 3 moles
- \( \text{Fe(NO}_3)_3 \): 2 moles
- Thus, the ratio becomes:
\[ \text{AgNO}_3 : \text{Pb(NO}_3)_2 : \text{Fe(NO}_3)_3 = 6 : 3 : 2 \]
- Simplifying this gives:
\[ \text{AgNO}_3 : \text{Pb(NO}_3)_2 : \text{Fe(NO}_3)_3 = 3 : 1.5 : 1 \]
- Further simplifying gives us:
\[ \text{AgNO}_3 : \text{Pb(NO}_3)_2 : \text{Fe(NO}_3)_3 = 6 : 3 : 2 \]

### Final Ratio:
The final ratio of the number of moles of \( \text{AgNO}_3 : \text{Pb(NO}_3)_2 : \text{Fe(NO}_3)_3 \) required for coagulation is:
\[ 6 : 3 : 2 \]