A solution containing 2.675 g of CoCl3 ⋅ 6NH3 (molar mass = 267 g mol-1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 g of AgCl. The formula of the complex is:

Question

A solution containing 2.675 g of CoCl3 ⋅ 6NH3 (molar mass = 267 g mol-1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 g of AgCl. The formula of the complex is: (A) [Co(NH3)6]Cl3 (B) [Co(NH3)3Cl3]3NH3 (C) [Co(NH3)4Cl2]Cl2NH3 (D) [Co(NH3)5Cl]Cl2NH3

Tardigrade Admin · Accepted Answer

No. of moles of complex =(2.675/267.5)=0.01 No. of moles of AgCl ==(4.78/143.5)=0.03 This shows that three Cl- ions are precipitated. Thus, formula of the complex is [Co(NH3)6]Cl3.

Q. A solution containing $2.675 g$ of $C o C l_{3} \cdot 6 N H_{3}$ (molar mass $= 267 g m o l^{- 1}$ ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $A g N O_{3}$ to give $4.78 g$ of $A g Cl$ . The formula of the complex is:

$[C o (N H_{3})_{6}] C l_{3}$

$[C o (N H_{3})_{3} C l_{3}]_{3} N H_{3}$

$[C o (N H_{3})_{4} C l_{2}] C l_{2} N H_{3}$

$[C o (N H_{3})_{5} Cl] C l_{2} N H_{3}$

No. of moles of complex = $\frac{2.675}{267.5} = 0.01$
No. of moles of AgCl == $\frac{4.78}{143.5} = 0.03$
This shows that three Cl- ions are precipitated. Thus, formula of the complex is $[C o (N H_{3})_{6}] C l_{3} .$

Q. A solution containing 2.675g of CoCl3​⋅6NH3​ (molar mass =267gmol−1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3​ to give 4.78g of AgCl. The formula of the complex is:

[Co(NH3​)6​]Cl3​

[Co(NH3​)3​Cl3​]3​NH3​

[Co(NH3​)4​Cl2​]Cl2​NH3​

[Co(NH3​)5​Cl]Cl2​NH3​