We have taken a saturated solution of `AgBr`, whose `K_(sp)` is `12xx10^(-14).` If `10^(-7)M` of `AgNO_(3)` are added to `1L` of this solutino, find the conductivity `(` specific conductance `)` of the solution in terms of `10^(-7)S m^(-1)` units.
Given `:`
`lambda^(@)._((Ag^(o+)))=6xx10^(-3)S m^(2) mol^(-1)`
`lambda^(@)._((Br^(c-)))=8xx10^(-3)S m^(2)mol^(-1)`
`lambda^(@)._((NO_(3)^(C-)))=7XX10^(-3)S m^(2) mol^(-1)`

Explanation

To solve the problem, we need to find the specific conductance (K) of a saturated solution of AgBr after adding AgNO3. The steps are as follows:

### Step 1: Determine the concentration of Ag⁺ and Br⁻ in the saturated solution of AgBr.
The solubility product constant (Ksp) for AgBr is given as \( K_{sp} = 12 \times 10^{-14} \).

For the dissociation of AgBr:
\[ \text{AgBr} \rightleftharpoons \text{Ag}^+ + \text{Br}^- \]

Let the solubility of AgBr be \( S \). Therefore, at equilibrium:
\[ [\text{Ag}^+] = S \]
\[ [\text{Br}^-] = S \]

From the Ksp expression:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] = S^2 \]
\[ 12 \times 10^{-14} = S^2 \]
\[ S = \sqrt{12 \times 10^{-14}} = \sqrt{1.2 \times 10^{-13}} \approx 3.46 \times 10^{-7} \, \text{M} \]

### Step 2: Calculate the total concentration of Ag⁺ after adding AgNO3.
When \( 10^{-7} \, \text{M} \) of AgNO3 is added, it contributes additional \( \text{Ag}^+ \) ions to the solution.

Total concentration of \( \text{Ag}^+ \):
\[ [\text{Ag}^+]_{total} = S + 10^{-7} = 3.46 \times 10^{-7} + 10^{-7} = 4.46 \times 10^{-7} \, \text{M} \]

### Step 3: The concentration of Br⁻ remains the same.
The concentration of Br⁻ remains equal to the solubility of AgBr:
\[ [\text{Br}^-] = S = 3.46 \times 10^{-7} \, \text{M} \]

### Step 4: Calculate the concentration of NO3⁻.
The concentration of \( \text{NO}_3^- \) from AgNO3 is:
\[ [\text{NO}_3^-] = 10^{-7} \, \text{M} \]

### Step 5: Calculate the individual conductivities.
Using the given molar conductivities:
- \( \lambda^0_{\text{Ag}^+} = 6 \times 10^{-3} \, \text{S m}^2 \text{mol}^{-1} \)
- \( \lambda^0_{\text{Br}^-} = 8 \times 10^{-3} \, \text{S m}^2 \text{mol}^{-1} \)
- \( \lambda^0_{\text{NO}_3^-} = 7 \times 10^{-3} \, \text{S m}^2 \text{mol}^{-1} \)

The total conductivity \( K \) is given by:
\[ K = \lambda^0_{\text{Ag}^+} [\text{Ag}^+] + \lambda^0_{\text{Br}^-} [\text{Br}^-] + \lambda^0_{\text{NO}_3^-} [\text{NO}_3^-] \]

Calculating each term:
1. For \( \text{Ag}^+ \):
\[ K_{\text{Ag}^+} = 6 \times 10^{-3} \times 4.46 \times 10^{-7} = 2.676 \times 10^{-9} \, \text{S m}^{-1} \]

2. For \( \text{Br}^- \):
\[ K_{\text{Br}^-} = 8 \times 10^{-3} \times 3.46 \times 10^{-7} = 2.768 \times 10^{-9} \, \text{S m}^{-1} \]

3. For \( \text{NO}_3^- \):
\[ K_{\text{NO}_3^-} = 7 \times 10^{-3} \times 10^{-7} = 7 \times 10^{-10} \, \text{S m}^{-1} \]

### Step 6: Sum the conductivities.
Now, summing these contributions:
\[ K_{total} = K_{\text{Ag}^+} + K_{\text{Br}^-} + K_{\text{NO}_3^-} \]
\[ K_{total} = 2.676 \times 10^{-9} + 2.768 \times 10^{-9} + 7 \times 10^{-10} \]
\[ K_{total} = 5.223 \times 10^{-9} \, \text{S m}^{-1} \]

### Step 7: Convert to the required units.
To express this in terms of \( 10^{-7} \, \text{S m}^{-1} \):
\[ K_{total} = 52.23 \times 10^{-7} \, \text{S m}^{-1} \]

Thus, the final answer is:
\[ \boxed{52.23} \]