Question

$AgCl(s) + Cl^{-} \rightleftharpoons AgCl_{2}^{-} $$$aM0$$

$\text{Final}\left( a - \frac{0.01}{100} \right)\text{ M1}\text{0}^{- 4}\text{ M}$

Now, $K_{eq} = \frac{\left\lbrack AgCl_{2}^{-} \right\rbrack}{\left\lbrack Cl^{-} \right\rbrack} \times \frac{\left\lbrack Ag^{+} \right\rbrack\left\lbrack Cl^{-} \right\rbrack}{\left\lbrack Ag^{+} \right\rbrack\left\lbrack Cl^{-} \right\rbrack} = K_{sp}.K_{f}$

$= 2 \times 10^{- 10} \times 2.5 \times 10^{5} = 5 \times 10^{- 5} $$$\therefore 5 \times 10^{- 5} = \frac{10^{- 4}}{\left( a - 10^{- 4} \right)} \Rightarrow a \approx 2\text{ M}$$

Hence, the minimum mass of NaCl added = (100 × 2) × 58.5 = 11700 gm