100 ml of 0.025M solution of chloride of ‘X’ required 200 ml of 0.05M solution of ${\mathrm{AgNO}}_3$ for complete precipitation. Formula of the chloride salt is

      Millimoles of ${\mathrm{Cl}}^{-}$ solution = 100(0.025) = 2.5
Millimoles of  ${\mathrm{AgNO}}_3$solution = 200(0.05) = 10
            Millimoles of  ${\mathrm{AgNO}}_3$ = ‘4’ times millimoles of chloride solution.
$\therefore \mathrm{Formula} \mathrm{of} \mathrm{the} \mathrm{salt} = {\mathrm{XCl}}_4$
${\mathrm{XCl}}_4+4{\mathrm{AgNO}}_3\to 4\mathrm{AgCl}+\mathrm{X}{\left({\mathrm{NO}}_3\right)}_4$