Answer
1.22 mol
Work Step by Step
$AgCl+2\ NH_3\leftrightarrow Ag(NH_3)_2^++Cl^-$
$K=Ksp\dot{}Kf=1.98\dot{}10^{-3}$
Full dissolution:
$[Ag(NH_3)_2^+]=[Cl^-]=0.05\ M$
$1.98\dot{}10^{-3}=0.05^2/[NH_3]^2\rightarrow [NH_3]=1.12\ M$
Total NH3 added:
$1.12\ M (NH_3)+ 2\dot{}0.05\ M (complex)=1.22\ M$
$n=1.22\ M\dot{}1.0\ L=1.22\ mol$