A solution containing 2.675 g of CoCl3.6NH3 (molar mass = 267.5 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 (molar mass = 143.5 g mol-1). The formula of the complex is (Atomic mass of Ag = 108 u)

Question

A solution containing 2.675 g of CoCl3.6NH3 (molar mass = 267.5 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 (molar mass = 143.5 g mol-1). The formula of the complex is (Atomic mass of Ag = 108 u) (A) [Co(NH3)6]Cl3 (B) [CoCl2(NH3)4]Cl (C) [CoCl3(NH3)3] (D) [CoCl(NH3)5]Cl2

Tardigrade Admin · Accepted Answer

Mole of CoCl3.6NH3=(2.675/267.5)=0.01 mol AgNO3 (aq)+Cl- (aq) → AgCl â (white) +NO-3 (aq) Mole of AgCl=(4.78/143.5)=0.03 mol 0.01 mol CoCl3. 6NH3 gives = 0.03 mol AgCl ∴ 1 mol CoCl3. 6NH3 ionises to give = 3 mol Cl- Hence, the formula of compound is [Co(NH3)6]Cl3

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

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

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

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

[Co(NH3​)6​]Cl3​

[CoCl2​(NH3​)4​]Cl

[CoCl3​(NH3​)3​]

[CoCl(NH3​)5​]Cl2​

Mole of CoCl3​.6NH3​=267.52.675​=0.01mol AgNO3​ (aq)+Cl− (aq) →AgCl↓ (white) +NO3−​ (aq) Mole of AgCl=143.54.78​=0.03mol 0.01molCoCl3​.6NH3​ gives =0.03molAgCl ∴1molCoCl3​.6NH3​ ionises to give = 3 mol Cl− Hence, the formula of compound is [Co(NH3​)6​]Cl3​

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

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

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

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