${Ksp}={IP}=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]=1.7 \times 10^{-10}$ $\left[{Ag}^{+}\right]=1.7 \times 10^{-9}$

$(ii)$ $\left[{Ag}^{+}\right]$required to ${ppt} {Ag}_{2} {CrO}_{4}({~s})$

${Ksp}={IP}=\left[{Ag}^{+}\right]^{2}\left[{CrO}_{4}^{-2}\right]=1.9 \times 10^{-12}$ $\left[{Ag}^{+}\right]=4.3 \times 10^{-5}$

$\left[{Ag}^{+}\right]$required to ${ppt} {AgCl}$ is low so ${AgCl}$ will ${ppt}$  ${1}^{{st}}$.