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

Formation of 0.2 mole of AgCl, from 0.1 mole of complex suggest the presence of two `Cl^(-)` outside the coordination sphere. Thus, the formula of the complex should be `[Co(NH_(3))_(5)Cl]Cl_(2)`. This complex ionises as
`[Co(NH_(3))_(5)Cl]Cl_(2) rarr underset(1)([Co(NH_(3))_(5)Cl]^(2+))+underset(2)(2Cl^(-))`
Hence, it is 1:2 electrolyte.