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 toA. `1 : 3` electrolyteB. `1 : 2` electrolyteC. `1 : 1` electrolyteD. `3 : 1` electrolyte

When 0.1 mol `CoCl_(3)(NH_(3))_(5)` is treated with excess of `AgNO_(3

1.	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 toA. `1 : 3` electrolyteB. `1 : 2` electrolyteC. `1 : 1` electrolyteD. `3 : 1` electrolyte
Answer» Correct Answer - B 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 .

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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 toA. `1 : 3` electrolyteB. `1 : 2` electrolyteC. `1 : 1` electrolyteD. `3 : 1` electrolyte

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