`2 L ` of `1 molal` solution of a complex salt `CrCl_(3).6H_(2)O (Mw=266.5)` shows an osmotic pressure of `98.52 atm`. The solution is now treated with `1 L` of `6 M AgNO_(3)`, which of the following are correct?

Options

A. Weight of `AgCl` precipitated is `861 g`.
B. The clear solution will show an osmotic pressure of `98.52 atm`.
C. The clear solution will show an osmotic pressure of `65.68 atm`.
D. `2 mol` of `[Cr(H_(2)O)_(6)] (NO_(3))_(3)` will be present in the solution.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the number of moles of the complex salt in the solution

Given:
- Volume of the solution = 2 L
- Molality of the solution = 1 molal

Using the definition of molality:
\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Assuming the mass of the solvent (water) is 1 kg (as is common in molality calculations), we can find the moles of solute:
\[ 1 = \frac{\text{moles of solute}}{1 \text{ kg}} \implies \text{moles of solute} = 1 \text{ mole} \]

### Step 2: Calculate the molarity of the solution

Molarity (M) is given by:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \]
Substituting the values:
\[ \text{Molarity} = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M} \]

### Step 3: Calculate the van 't Hoff factor (i) using osmotic pressure

The formula for osmotic pressure (\(\pi\)) is given by:
\[ \pi = iCRT \]
Where:
- \(\pi = 98.52 \text{ atm}\)
- \(C = 0.5 \text{ M}\)
- \(R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\)
- \(T = 298 \text{ K}\) (assuming room temperature)

Rearranging the formula to find \(i\):
\[ i = \frac{\pi}{CRT} \]
Substituting the known values:
\[ i = \frac{98.52}{0.5 \times 0.0821 \times 298} \]
Calculating the denominator:
\[ 0.5 \times 0.0821 \times 298 \approx 12.2 \]
Thus:
\[ i = \frac{98.52}{12.2} \approx 8.08 \]

### Step 4: Analyze the dissociation of the complex salt

The complex salt \(CrCl_3 \cdot 6H_2O\) dissociates in solution. The dissociation can be represented as:
\[ CrCl_3 \cdot 6H_2O \rightarrow Cr^{3+} + 3Cl^{-} + 6H_2O \]
This gives a total of \(1 + 3 = 4\) ions from the salt. However, the calculated \(i\) value suggests that there are additional ions, likely due to the hydration of the complex.

### Step 5: Reaction with AgNO3

When treated with \(1 \text{ L}\) of \(6 \text{ M} \, AgNO_3\):
- Moles of \(AgNO_3 = 6 \text{ moles}\)
- The \(CrCl_3\) will react with \(AgNO_3\) to form \(AgCl\):
\[ 3Cl^{-} + 3AgNO_3 \rightarrow 3AgCl + 3NO_3^{-} \]
Since we have 1 mole of \(CrCl_3\), it will produce 3 moles of \(AgCl\).

### Step 6: Calculate the mass of \(AgCl\) formed

The molar mass of \(AgCl\) is approximately \(143.5 \text{ g/mol}\):
\[ \text{Mass of } AgCl = 3 \text{ moles} \times 143.5 \text{ g/mol} = 430.5 \text{ g} \]

### Conclusion

The correct statements regarding the solution after the reaction with \(AgNO_3\) are:
1. 430.5 g of \(AgCl\) is formed.
2. The osmotic pressure will not remain the same after the reaction.