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Q.

When 0.1 molCoCl3(NH3)5 is treated with excess ofAgNO3, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to

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a

1:3 electrolyte

b

1:2 electrolyte

c

1:1 electrolyte

d

3:1 electrolyte

answer is B.

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Detailed Solution

When 0.1 molCoCl3(NH3)5 is treated with excess ofAgNO3, 0.2 mol of AgCl are obtained. [CoCl(NH3)5]Cl2→[CoCl(NH3)5]+2+2Cl−1Hence, conductivity of solution correspond to 1: 2.

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