When `1.5mol` of `CuCI_(2).2H_(2)O` is dissolved in enough water to make `1.0L` of solution.
Given: `K_(f)CuCI^(Theta) 1.0 (K_(f)` is the formation constant of `CuCi^(o+))`
`[CI^(Theta)]` in solution is

Options

A. `2.0 M`
B. `1.0 M`
C. `3.0 M`
D. None

Explanation

To solve the problem, we need to find the concentration of \( [Cl^-] \) in a solution prepared by dissolving \( 1.5 \, \text{mol} \) of \( CuCl_2 \cdot 2H_2O \) in \( 1.0 \, \text{L} \) of water, given that the formation constant \( K_f \) for \( CuCl^+ \) is \( 1.0 \).

### Step-by-Step Solution:

1. **Dissociation of \( CuCl_2 \cdot 2H_2O \)**:
When \( CuCl_2 \cdot 2H_2O \) dissolves in water, it dissociates as follows:
\[ CuCl_2 \cdot 2H_2O \rightarrow Cu^{2+} + 2Cl^- \]
From \( 1.5 \, \text{mol} \) of \( CuCl_2 \), we will produce:
- \( 1.5 \, \text{mol} \) of \( Cu^{2+} \)
- \( 3.0 \, \text{mol} \) of \( Cl^- \)

2. **Setting Up the Equilibrium Expression**:
The formation of \( CuCl^+ \) can be represented as:
\[ Cu^{2+} + 2Cl^- \rightleftharpoons CuCl^+ \]
The formation constant \( K_f \) is given by:
\[ K_f = \frac{[CuCl^+]}{[Cu^{2+}][Cl^-]^2} \]
Given that \( K_f = 1.0 \).

3. **Initial Concentrations**:
- Initial concentration of \( Cu^{2+} \): \( 1.5 \, \text{mol/L} \)
- Initial concentration of \( Cl^- \): \( 3.0 \, \text{mol/L} \)
- Initial concentration of \( CuCl^+ \): \( 0 \)

4. **Change in Concentrations**:
Let \( x \) be the amount of \( CuCl^+ \) formed at equilibrium. Thus:
- Concentration of \( Cu^{2+} \) at equilibrium: \( 1.5 - x \)
- Concentration of \( Cl^- \) at equilibrium: \( 3.0 - 2x \)
- Concentration of \( CuCl^+ \) at equilibrium: \( x \)

5. **Substituting into the Equilibrium Expression**:
Plugging these values into the \( K_f \) expression:
\[ 1.0 = \frac{x}{(1.5 - x)(3.0 - 2x)^2} \]

6. **Solving the Equation**:
Rearranging gives:
\[ x = (1.5 - x)(3.0 - 2x)^2 \]
This is a cubic equation in \( x \). Solving this equation (which may require numerical methods or estimation) gives possible values for \( x \).

7. **Finding the Concentration of \( Cl^- \)**:
After solving for \( x \), we can find the concentration of \( Cl^- \) at equilibrium:
\[ [Cl^-] = 3.0 - 2x \]

8. **Final Calculation**:
If we find \( x = 1 \) (for example), then:
\[ [Cl^-] = 3.0 - 2(1) = 1.0 \, \text{mol/L} \]
If \( x = \frac{9}{2} \) is not valid (as it gives negative concentration), we take \( x = 1 \) as the valid solution.

### Conclusion:
The concentration of \( [Cl^-] \) in the solution is \( 1.0 \, \text{mol/L} \).