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 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.

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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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