A solution containing 2.675 of CoCl3. 6NH3(molar mass = 267.5 g mol–1) is passed through a cation exchanger. The chloride ions obtained is solution was treated with excess of AgNO3 to give 4.78 g of AgCl (molar mass = 143.5 g mol–1 ). The formula of the complex is
(At. mass of Ag = 108 u)
-
[Co(NH3)6]Cl3
-
[CoCl2(NH3)4]Cl
-
[CoCl3(NH3)3]
-
[CoCl(NH3)5]Cl2
A.
[Co(NH3)6]Cl3
CoCl. 6NH3 + AgNO3 → AgCl
2.675/2667.5 excess 4.78/143.5
= 0.01 mole = 0.03310 mole
because 0.01 mole CoCl3.6NH3 given 0.0331 mole AgCl.
hence 1 mole of CoCl3.6NH3 will given 0.03310/0.010 =3 mole.
Thus the formula of the compound will be [Co(NH3)6]Cl3