1.

When 0.1 mol CoCl_(3) (NH_(3))_(5) is treated with excess of AgNO_(3), 0.1 mol of AgCl are obtained. The conductivity of solution will correspond to

Answer»

1 : 3 electrolyte
1 : 2 electrolyte
1 : 1 electrolyte
3 : 1 electrolyte

Solution :FORMATION of 0.2 MOL of AgCl from 0.1 mol of the complex means that there are two ionizable CL. Hence, formula is `[Co(NH_(3))_(5)Cl]Cl_(2)`, i.e., 1 : 2 type electrolyte.


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