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When 0.1 mol CoCl(3)(NH(3))(5) is treate...

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 correspound to -

A

`1:3` electrolyte

B

`1:2` electrolyte

C

`1:1` electrolyte

D

`3:1` electrolyte

Text Solution

Verified by Experts

The correct Answer is:
B
Precipitation of 02 mol of AgCl from 0.1 mol of complex indicates that each molecule of the complex contains two insoluble Cl atoms. So, the complex is `[CoCl(NH_(3))_(5)]Cl_(2)`, which corresponds to `1:2` electrolyte.
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