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

When 0.1 mole of `CoCl_(3)(NH_(3))_(5)` was reacted with excess of `AgNO_(3)`, we get 0.2 moles of AgCl. So, there are two chloride ions that are free and not part of the complex. The formula for complex has to be [`Co(NH_(3))_(5)Cl]Cl_(2)`.
`[Co(NH_(3))Cl]Cl_(2) to [Co(NH_(3))_(5)Cl]^(2-)+2Cl^(-)`
Therefore, the conductivity of the solution will be 1:2 electrolyte.