When 0.1 mol CoCl3(NH3)5 is treated with excess of AgNO3, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to

Question

When 0.1 mol CoCl3(NH3)5 is treated with excess of AgNO3, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to (A) 1: 3 electrolyte (B) 1: 2 electrolyte (C) 1: 1 electrolyte (D) 3: 1 electrolyte

Tardigrade Admin · Accepted Answer

0.2 mol of AgCl are obtained when 0.1 mol CoCl3(NH3)5 is treated with excess of AgNO3 which shows that one molecule of the complex gives two Cl- ions in solution. Thus, the formula of the complex is [Co(NH3)5Cl]Cl2 i.e., 1: 2 electrolyte.

Q. When $0.1 m o l C o C l_{3} (N H_{3})_{5}$ is treated with excess of $A g N O_{3}$ , $0.2$ mol of $A g Cl$ are obtained. The conductivity of solution will correspond to

$1 : 3$ electrolyte

$1 : 2$ electrolyte

$1 : 1$ electrolyte

$3 : 1$ electrolyte

$0.2$ mol of $A g Cl$ are obtained when $0.1$ mol $C o C l_{3} (N H_{3})_{5}$ is treated with excess of $A g N O_{3}$ which shows that one molecule of the complex gives two $C l^{-}$ ions in solution. Thus, the formula of the complex is $[C o (N H_{3})_{5} Cl] C l_{2}$ i.e., $1 : 2$ electrolyte.

Q. When 0.1molCoCl3​(NH3​)5​ is treated with excess of AgNO3​, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to

1:3 electrolyte

1:2 electrolyte

1:1 electrolyte

3:1 electrolyte