The conductivity `(k)` of a saturated solution of `AgBr` at `298K` is `8.5xx10^(-7)Scm^(-1)`. If `lambda^(@)._(Ag^(o+))` and `lambda^(@)._(Br^(c-))` are 62 and `78 S cm^(2)mol^(-1)` , respectively, then calculate the solubility and `K_(sp)` of `AgBr`.

Explanation

To solve the problem, we need to calculate the solubility and the solubility product constant (Ksp) of AgBr based on the given conductivity and molar conductivities. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Conductivity of saturated AgBr solution, \( k = 8.5 \times 10^{-7} \, \text{S cm}^{-1} \)
- Molar conductivity of \( \text{Ag}^+ \), \( \lambda^0_{\text{Ag}^+} = 62 \, \text{S cm}^2 \text{mol}^{-1} \)
- Molar conductivity of \( \text{Br}^- \), \( \lambda^0_{\text{Br}^-} = 78 \, \text{S cm}^2 \text{mol}^{-1} \)

### Step 2: Calculate the molar conductivity of AgBr
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