Explanation

(a)
Given, concentration of $NaCl = 0.2\ M$
$K_{sp}(AgCl\ ) = 1.20 \times 10^{- 10}$
Let the solubility of $AgCl\ in\ NaCl = x$
$AgCl\ \rightarrow Ag^{+} + Cl^{-}$
$x$ $x$ $x$
Solubility $NaCl \rightarrow {Na}^{+} + {Cl}^{-}$
0.2 0.2 0.2
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrack Ag^{+} \right\rbrack = x\ and\ \left\lbrack Cl^{-} \right\rbrack = (x + 0.2)$
$\therefore\ \ \ \ \ \ \ \ K_{sp}(AgCl\ ) = \left\lbrack Ag^{+} \right\rbrack\left\lbrack Cl^{-} \right\rbrack\$
$= x(x + 0.2)$
$= x^{2} + 0.2x$
$\therefore\ \ \ \ \ \ \ K_{sp} = 0.2x(x^{2} \ll 1)$
or $1.2 \times 10^{- 10} = 0.2x$
$\therefore$ $x = 6 \times 10^{- 10}$