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  • When 0.1 mol {CoCl}_{3}({NH}_{3})_{5} is treated with an excess of {AgNO}_{3}, 0.2 mol of {AgCl} are obtained. The conductivity of the solution will correspond to (i) 1:3 electrolyte (ii) 1:2 electrolyte (iii) 1:1 electrolyte

When 0.1 mol \text {CoCl}_{3}\left ( \text {NH}_{3} \right )_{5} is treated with an excess of \text {AgNO}_{3}, 0.2 mol of \text {AgCl} are obtained. The conductivity of the solution will correspond to

(i) 1:3 electrolyte

(ii) 1:2 electrolyte

(iii) 1:1 electrolyte

(iv) 3:1 electrolyte

Answers (1)

Option (ii) is the correct answer.

Explanation:

One mole of chloride ion gets precipitated by one mole of \text {AgNO}_{3}. \text {AgCl}, is obtained when 0.1 mole of \text {CoCl}_{3}\left ( \text {NH}_{3} \right )_{5} is treated with \text {AgNO}_{3} in excess. So this leaves two free chloride ions in the solution of electrolyte for every 1 mole reaction.

Hence the molecular formula should be \left [ \text {Co}\left ( \text {NH}_{3} \right )_{5}\text {Cl} \right ]\text {Cl}_{2} and the solution of electrolyte should contain \left [ \text {Co}\left ( \text {NH}_{3} \right )_{5}\text {Cl} \right ]^{2+} and two Cl as their constituent ions. Therefore, it is an 1:2 electrolyte.

\left [ \text {Co}\left ( \text {NH}_{3} \right )_{5}\text {Cl} \right ]\text {Cl}_{2} \rightarrow \left [ \text {Co}\left ( \text {NH}_{3} \right )_{5}\text {Cl} \right ]^{2+} \text {(aq)}+\text {2Cl}^{-} \text {(aq)}

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