1.0 molar solution of the complex of the salt, `CrCl_(3).6H_(2)O`, displays an osmotic pressure of 3RT. 0.5 L of the same solution on treatment with excess of `AgNO_(3)` solution will yield (assume `alpha` = 1) :

Options

A. 0.5 mole of AgCl
B. 1.0 mole of AgCl
C. 1.5 mole of AgCl
D. 3.0 mole of AgCl

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript while ensuring clarity and completeness in each step.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- We have a 1.0 M solution of the complex salt `CrCl3.6H2O`.
- The osmotic pressure (π) of this solution is given as `3RT`.
- We are to find out how much `AgCl` will be produced when 0.5 L of this solution is treated with excess `AgNO3`.

2. **Use the Osmotic Pressure Formula:**
- The formula for osmotic pressure is given by:
\[ \pi = iCRT \]
- Here, \(C\) (concentration) is 1.0 M, and we know that \(\pi = 3RT\).

3. **Calculate the Van 't Hoff Factor (i):**
- From the osmotic pressure equation:
\[ 3RT = i \cdot 1 \cdot RT \]
- We can cancel \(RT\) from both sides:
\[ 3 = i \]
- Therefore, the van 't Hoff factor \(i = 3\).

4. **Relate i to the Degree of Dissociation (α) and Number of Ions (n):**
- The relationship is given by:
\[ i = 1 + (n - 1) \alpha \]
- Given that \(\alpha = 1\) (complete dissociation), we can substitute:
\[ 3 = 1 + (n - 1) \cdot 1 \]
- Simplifying this gives:
\[ 3 = 1 + n - 1 \implies n = 3 \]

5. **Identify the Dissociation of the Complex:**
- The complex `CrCl3.6H2O` dissociates into:
\[ Cr^{3+} + 3Cl^{-} \]
- This means that 1 mole of the complex yields 1 mole of `Cr^{3+}` and 3 moles of `Cl^{-}`.

6. **Determine the Reaction with `AgNO3`:**
- When `AgNO3` is added, `Cl^{-}` ions will react to form `AgCl`:
\[ Cl^{-} + AgNO3 \rightarrow AgCl + NO3^{-} \]
- Since there are 3 moles of `Cl^{-}` from 1 mole of the complex, it will produce 3 moles of `AgCl`.

7. **Calculate the Amount of `AgCl` from 0.5 L of Solution:**
- Since the concentration of the solution is 1.0 M, 0.5 L of the solution contains:
\[ 0.5 \text{ L} \times 1.0 \text{ mol/L} = 0.5 \text{ moles of complex} \]
- Each mole of complex produces 3 moles of `AgCl`, so:
\[ 0.5 \text{ moles of complex} \times 3 \text{ moles of } AgCl/\text{mole of complex} = 1.5 \text{ moles of } AgCl \]

### Final Answer:
The treatment of 0.5 L of the solution with excess `AgNO3` will yield **1.5 moles of `AgCl`**.