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

0.2 mole of AgCl are obtained when 0.1 mol `CoCl_(3) (NH_(3))_(5)` is treated with excess of `AgNO_(3)` which show that one molecule of the complex gives two `Cl^(-)`ions in solution . Thus , the formula of the complex is `[Co(NH_(3))_(5) Cl]Cl_(2)` i.e., `1 : 2` electrolyte .