Question

When 0.1 mol of cocl3 (nh3)5 is treated with excess of agno3 , 0.2 mol of agcl is obtained . The conductivity of solution will corresponds to?

Answer 1

Answer:

1)One mole of $AgNO_{3}$ precipitates one mole of chloride ion.

In the above reaction when 0.1 mole $COCl_{3} (NH_{3} )_{5}$ is treated with excess of $AgNO_{3}$

2)0.2 moles of AgCl are obtained thus there must be two free chloride ions in the solution of electrolyte.

so electrolyte is in the ratio of 1:2

[$Co(NH_{3} )_{5} Cl]Cl_{2}$→[$Co(NH_{3} )_{5} Cl]^{2+} (aq)+2Cl^{-} (aq)$