Minimum moles of NH _ 3 required to be added to 1L solution so as...

$\begin{cases}\text{AgCl} \left(\text{s}\right) & + & \text{aq} & ⇌ & {Ag}_{\text{aq}}^{+} & + & {Cl}_{aq}^{-} & \text{;} & {K}_{sp} \\ {Ag}_{aq}^{+} & + & 2 {NH}_{3 \text{aq}} & ⇌ & \left[\text{Ag} {\left({NH}_{3}\right)}_{2 \text{aq}}^{+}\right] & & & \text{;} & {K}_{f}\end{cases}$

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$\begin{cases}\text{AgCl} \left(\text{s}\right) & + & 2 {NH}_{3} & ⇌ & \left[\text{Ag} {\left({NH}_{3}\right)}_{2 \text{aq}}^{+}\right] & + & {Cl}_{aq}^{-} & & \text{K} = {K}_{sp} & \times & {K}_{f}\end{cases}$

~ & \left(C - 0 .2\right)M0 .1 M0 .1 M$Equilibrium concentration<br /><br />$\text{K} = 1 {0}^{- 1 0} \times 1 {0}^{8} = 1 {0}^{- 2} = \dfrac{\left[\text{Ag} [ {NH}_{3} {)}_{2}^{+}\right] ~\left[{Cl}^{-}\right]}{{\left[{NH}_{3}\right]}^{2}}$<br /><br />$1 {0}^{- 2} = \dfrac{1 {0}^{- 1} \times 1 {0}^{- 1}}{{\left(\text{M} - 0 \text{.} 2\right)}^{2}}$<br /><br />$∴ M - 0 .2 = 1$<br /><br />$∴ M = 1 .2 M