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

Options

A. 1 : 3 electrolyte
B. 1 : 2 electrolyte
C. 1 : 1 electrolyte
D. 3 : 1 electrolyte

Answer

B

Explanation

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)`, 0.2 mole of `AgCl` are obtained thus, these must be two free chloride ios in the solution of electrolyte.
So, molecular formula of complex will be `[Co(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) rarr [Co(NH_(3))_(5)Cl]^(2 o+) (aq) + 2Cl^(-) (aq)`
Hence, option (b) is the correct.