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 (Ksp) of AgCl, we will follow these steps:

### Step 1: Calculate Molar Conductance at Infinite Dilution of AgCl
The molar conductance at infinite dilution (λM^0) of AgCl can be calculated using the formula:

\[ λM^0_{AgCl} = λM^0_{AgNO_3} + λM^0_{NaCl} - λM^0_{NaNO_3} \]
...