A saturated solution in `AgX(K_(sp)=3xx10^(-12))` and `AgY(K_(sp)=10^(-12))` has conductivity `0.4xx10^(-6)Omega^(-1)cm^(-1)`.
Given: Limiting molar conductivity of `Ag^+=60Omega^(-1)cm^2mol^(-1)`
Limiting molar conductivity of `X^(-)=90Omega^(-1)cm^2mol^(-1)`
The conductivity of `Y^(-)` is (in `Omega^(-1)cm^(-1)`):

Options

A. `1.45xx10^(-7)`
B. `1.45xx10^(-5)`
C. `1.45xx10^(-9)`
D. None of these

Answer

To solve the problem, we need to find the conductivity of the anion \( Y^- \) in a saturated solution of \( AgX \) and \( AgY \). We are given the solubility products \( K_{sp} \) for both salts, the conductivity of the solution, and the limiting molar conductivities of \( Ag^+ \) and \( X^- \).

### Step-by-Step Solution:

1. **Understanding the Dissociation of Salts:**
- For \( AgX \):
\[ AgX \rightleftharpoons Ag^+ + X^- \]
Let the solubility of \( AgX \) be \( s \). Then, at equilibrium:
\[ [Ag^+] = s \quad \text{and} \quad [X^-] = s \]
Thus, the solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [Ag^+][X^-] = s \cdot s = s^2 \]
Given \( K_{sp} = 3 \times 10^{-12} \):
\[ s^2 = 3 \times 10^{-12} \implies s = \sqrt{3 \times 10^{-12}} = \sqrt{3} \times 10^{-6} \approx 1.73 \times 10^{-6} \, \text{mol/L} \]

2. **For \( AgY \):**
- For \( AgY \):
\[ AgY \rightleftharpoons Ag^+ + Y^- \]
Let the solubility of \( AgY \) be \( y \). Then:
\[ [Ag^+] = y \quad \text{and} \quad [Y^-] = y \]
The solubility product \( K_{sp} \) is:
\[ K_{sp} = [Ag^+][Y^-] = y \cdot y = y^2 \]
Given \( K_{sp} = 10^{-12} \):
\[ y^2 = 10^{-12} \implies y = \sqrt{10^{-12}} = 10^{-6} \, \text{mol/L} \]

3. **Total Concentration of Ions:**
- The total concentration of \( Ag^+ \) ions from both salts is:
\[ [Ag^+] = s + y = 1.73 \times 10^{-6} + 10^{-6} = 2.73 \times 10^{-6} \, \text{mol/L} \]

4. **Calculating Conductivity:**
- The conductivity \( \kappa \) of the solution is given by:
\[ \kappa = c \cdot (\lambda_{Ag^+} + \lambda_{X^-} + \lambda_{Y^-}) \]
Where \( c \) is the concentration of ions. We know:
\[ \kappa = 0.4 \times 10^{-6} \, \Omega^{-1} \text{cm}^{-1} \]
Given:
- \( \lambda_{Ag^+} = 60 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
- \( \lambda_{X^-} = 90 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)

5. **Setting Up the Equation:**
- The equation for conductivity becomes:
\[ 0.4 \times 10^{-6} = (2.73 \times 10^{-6}) \cdot (60 + 90 + \lambda_{Y^-}) \]
- Simplifying:
\[ 0.4 \times 10^{-6} = (2.73 \times 10^{-6}) \cdot (150 + \lambda_{Y^-}) \]

6. **Solving for \( \lambda_{Y^-} \):**
- Rearranging gives:
\[ 0.4 \times 10^{-6} = 2.73 \times 10^{-6} \cdot (150 + \lambda_{Y^-}) \]
\[ \frac{0.4 \times 10^{-6}}{2.73 \times 10^{-6}} = 150 + \lambda_{Y^-} \]
\[ \lambda_{Y^-} = \frac{0.4}{2.73} - 150 \]
- Calculating:
\[ \lambda_{Y^-} \approx 1.47 \times 10^{-7} \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Final Answer:
The conductivity of \( Y^- \) is approximately \( 1.45 \times 10^{-7} \, \Omega^{-1} \text{cm}^{-1} \).