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InterviewSolution
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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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Answer» Correct Answer - B One mole of `AgNO_(3)` precipitates one mole of chloride ion. In the above reaction, when 0.1 mole `CoCl_(3) (NH_(3))_(5)` is treated with excess of `AgNO_(3)`, 0.2 mole of `AgCl` are obtained thus, these must be two free chloride ios in the solution of electrolyte. So, molecular formula of complex will be `[Co(NH_(3))_(5) Cl] Cl_(2)`and electrolytic solution must contain `[Co (NH_(3))_(5)Cl]^(2+)` and two `Cl^(-)` as constituent ions. Thus, it is 1 : 2 electrolyte. `[Co(NH_(3))_(5)Cl]Cl_(2) rarr [Co(NH_(3))_(5)Cl]^(2 o+) (aq) + 2Cl^(-) (aq)` Hence, option (b) is the correct. |
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