On treatment of 10 ml of 1M solution of the complex `CrCl_(3).6H_(2)O` with excess of `AgNO_(3)`, 4.305 g of AgCl was obtained. The complex is

Options

A. `[Cr(H_(2)O)_(3)Cl_(3)].3H_(2)O`
B. `[Cr(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O`
C. `[Cr(H_(2)O)_(5)Cl)Cl_(2).H_(2)O`
D. `[Cr(H_(2)O)_(6)]Cl_(3)`

Answer

To solve the problem, we need to determine the nature of the complex `CrCl3.6H2O` based on the reaction with `AgNO3` that produces `AgCl`. Let's break down the solution step by step.

### Step 1: Calculate the number of moles of the complex

We are given:
- Volume of the solution = 10 mL = 10/1000 L = 0.01 L
- Molarity of the solution = 1 M

The number of moles of the complex can be calculated using the formula:

\[ \text{Number of moles} = \text{Volume (L)} \times \text{Molarity (mol/L)} \]

Substituting the values:

\[ \text{Number of moles} = 0.01 \, \text{L} \times 1 \, \text{mol/L} = 0.01 \, \text{moles} \]

### Step 2: Calculate the number of moles of AgCl produced

We are given:
- Mass of AgCl obtained = 4.305 g
- Molar mass of AgCl = 143.32 g/mol

The number of moles of AgCl can be calculated using the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Substituting the values:

\[ \text{Number of moles of AgCl} = \frac{4.305 \, \text{g}}{143.32 \, \text{g/mol}} \approx 0.0300 \, \text{moles} \]

### Step 3: Determine the stoichiometry of the reaction

From the calculations, we have:
- 0.01 moles of the complex produces 0.0300 moles of AgCl.

This indicates that:
\[ 1 \, \text{mole of complex} \rightarrow 3 \, \text{moles of AgCl} \]

### Step 4: Identify the complex

The complex given is `CrCl3.6H2O`. When this complex dissociates in solution, it can be represented as:

\[ \text{Cr}^{3+} + 3 \text{Cl}^- \]

This means that for every mole of `CrCl3`, three moles of chloride ions are released, which can react with `AgNO3` to form `AgCl`.

### Conclusion

Since we need 3 moles of AgCl for every mole of the complex, the correct representation of the complex is `Cr(H2O)6Cl3`. This matches with the stoichiometry we derived from the reaction.

### Final Answer

The complex is `Cr(H2O)6Cl3`.

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