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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 mole 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

Text Solution

Verified by Experts

The correct Answer is:
B
`[Co(NH_(3))_(5)Cl]Cl_(2)implies1:2` electrolyte
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