One mole of $\mathrm{AgCl}$ obtained as precipitate by treating with ____ mol of aqueous ${\mathrm{CrCl}}_3.5{\mathrm{NH}}_3$ with excess aqueous silver nitrate.

Reaction will be 

$[ \mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl} ] {\mathrm{Cl}}_2$is the complex one mole of which can give two moles of ppt and therefore half mole of complex will be enough to give one mole of ppt

The required mole of ${\mathrm{CrCl}}_3.5{\mathrm{NH}}_3$ is $0.50$.