In 1 L saturated solution of AgCl[ K _ sp (AgCl) = 1.6 × 1 0 ^ - 10...

Let the solubility of $AgCl$ is $x$ mol $\text{litr}{\text{e}}^{- 1}$

$\text{AgCl} ⇌ \mathop{\text{A}{\text{g}}^{+}}_{x} + \mathop{\text{C}{\text{l}}^{-}}_{x}$

And that of $CuCl$ is $y$ mol $\text{litr}{\text{e}}^{- 1}$

$\text{CuCl} ⇌ \mathop{\text{C}{\text{u}}^{+}}_{y} + \mathop{\text{C}{\text{l}}^{-}}_{y}$

${K}_{\text{sp}}$ of $\text{AgCl} = \left[\text{A}{\text{g}}^{+}\right]\left[\text{C}{\text{l}}^{- 1}\right]$

$1.6 \times 1{0}^{- 10} = x\left(x + y\right)$ ...(i)

Similarly ${K}_{\text{sp}}$ of $\text{CuCl} = \left[\text{C}{\text{u}}^{+}\right]\left[\text{C}{\text{l}}^{-}\right]$

$1.6 \times 1{0}^{- 6} = y\left(x + y\right)$ ...(ii)

On solving (i) and (ii)

\left[\text{A}{\text{g}}^{+}\right] = 1.6 \times 1{0}^{- 7}$<br/><br/>$x = 7