When 0.1 mol CoCl 3 (NH 3 ) 5 is treated with excess of AgNO 3 , 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to

Question

Philoid · Accepted Answer

When 0.1 mole of CoCl₃(NH₃)₅ was reacted with excess of AgNO₃, we get 0.2 moles of AgCl. So, there are two chloride ions that are free and not part of the complex. The formula for complex has to be [Co(NH₃)₅Cl]Cl₂.

[Co(NH₃)₅Cl]Cl₂→[Co(NH₃)₅Cl]⁺ + 2Cl^-

Therefore, the conductivity of the solution will be 1:2 electrolyte

[Co(NH₃)₅Cl]Cl₂→[Co(NH₃)₅Cl]⁺ + 2Cl^-

Therefore, the conductivity of the solution will be 1:2 electrolyte

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

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