A solution is $0.1\mathrm{M}$ in ${\mathrm{Cl}}^{-}$ and $0.001\mathrm{M}$ in ${\mathrm{CrO}}_4^{2-}.$
Solid ${\mathrm{AgNO}}_3$ is gradually added to it Assuming that the addition does not change in volume and ${\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{AgCl}\right)=1.7\times {10}^{-10}{\mathrm{M}}^2$ and
${\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)=1.9\times {10}^{-12}{\mathrm{M}}^3.$

Select correct statement from the following:

• A

  $\mathrm{AgCl}$ precipitates first because its ${\mathrm{K}}_{\mathrm{sp}}$ is high.
• B

  ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$ precipitates first as its ${\mathrm{K}}_{\mathrm{sp}}$ is low.
• C

  ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$ precipitates first because the amount of ${\mathrm{Ag}}^{+}$ needed is low.
• Correct answer
  D

  $\mathrm{AgCl}$ will precipitate first as the amount of ${\mathrm{Ag}}^{+}$ needed to precipitate is low.

Hence, the correct answer is option: 'D'.