If conductivity of water used to make saturated of AgCl is found to be `3.1xx10^(-5)Omega^(-1)`
`cm^(-1)` and conductance of the solution of `AgCl=4.5xx10^(-5)Omega^(-1)cm^(-1)`
if `lamda_(M)^(0)AgNO_(3)=200Omega^(-1)cm^(2)` `"mole"^(-1)`
`lamda_(M)^(0)NaNO_(3)=310Omega^(-1)cm^(2)` `"mole"^(-1)`
calculate `K_(SP)` of `AgCl`

Explanation

To calculate the solubility product constant (Ksp) of AgCl using the provided data, we will follow these steps:

### Step 1: Calculate the Molar Conductance of AgCl
The molar conductance at infinite dilution (λM) for AgCl can be calculated using the formula:

\[ \lambda_{M}^{0} (AgCl) = \lambda_{M}^{0} (AgNO_3) + \lambda_{M}^{0} (NaCl) - \lambda_{M}^{0} (NaNO_3) \]
...