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Karnataka Board PUCPUC Science 2nd PUC Class 12

When 0.1 mol CoClX3 (NHX3)X5 is treated with excess of AgNOX3, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to ______.

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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

MCQ
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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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Chapter 9: Coordination Compounds - Exercises [Page 120]

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NCERT Exemplar Chemistry Exemplar [English] Class 12
Chapter 9 Coordination Compounds
Exercises | Q I. 3. | Page 120

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