Question

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

Answer 1

Correct Answer - 2
One mole of `AgNO_(3)` precipitates one mole of chloride ion. In the given reaction, when 0.1 mole `CoCl_(3)(NH_(3))_(5)Cl]Cl_(2)` and electrolytic solution must contain `[Co(NH_(3))_(5)Cl]^(2+)` and two `Cl^(-)` as constituent ions. Thus it is 1:2 electrolyte.
`[Co(NH_(3))_(5)Cl]Cl_(2) to [Co(NH_(3))_(5)Cl]_(aq)^(2+) + 2Cl_(aq)^(-)` Hence, option (2) is the correct