`14.625mg` of `NaCl` is treated with `AgNO_(3)(aq)` .The mass of `AgCl` obtained as ppt is (`Ag=108g//mol`)

5.33 mg of salt [Cr(H_(2)O)_(5)Cl].Cl_(2) H_(2)O is treated with excess of AgNO_(3)(aq) then mass of AgCl ppt. obtained will be : [Given :Cr=52, Cl = 35.5]

When 1 mol CrCl_(3).6H_(2)O is treated with excess of AgNO_(3) , 3 mol of AgCl are obtained. The formula of the coplex is

When 0.1 mol CoCl_(3)(NH_(3))_(5) is treated with excess of AgNO_(3) , 0.2 mole of AgCl are obtained. The conductivity of solution will correspond to