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Question
When 0.1 mol \[\ce{CoCl3 (NH3)5}\] is treated with excess of \[\ce{AgNO3}\], 0.2 mol of \[\ce{AgCl}\] are obtained. The conductivity of solution will correspond to ______.
Options
1:3 electrolyte
1:2 electrolyte
1:1 electrolyte
3:1 electrolyte
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Solution
When 0.1 mol \[\ce{CoCl3 (NH3)5}\] is treated with excess of \[\ce{AgNO3}\], 0.2 mol of \[\ce{AgCl}\] are obtained. The conductivity of solution will correspond to 1:2 electrolyte.
Explanation:
One mole of \[\ce{AgNO3}\] precipitates one mole of chloride ion. In the above reaction, when 0.1 mole \[\ce{CoCl3 (NH3)5}\] is treated with excess of \[\ce{AgNO3}\], 0.2 mol of \[\ce{AgCl}\] are obtained thus, there must be two free chloride ions in the solution of electrolyte.
So, molecular formula of complex will be \[\ce{[CO(NH3)5Cl]Cl2}\] and electrolytic solution must contain \[\ce{[CO(NH3)5Cl]^{2+}}\] and two CT as constituent ions. Thus, it is 1:2 electrolyte.
\[\ce{[Co(NH3)5Cl]Cl2 -> [Co(NH3)5Cl]^{2+} (aq) + 2Cl^{-} (aq)}\]
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