1.

When 0.1 mol of CoCl_(3)(NH_(3))_(5) is treated with excess of AgNO_(3), 0.2 mol ofAgcl is obtained. The conductivity of solution will correspond to

Answer»

1:3 electroylte
1:2 electroyle
1:1 electrolyte
3:1 electrolyte

Solution :0.2 mol of Agcl is obtained when 0.1 mol of `CoCl_(3)(NH_(3))_(5)` is treatd with EXCESS of `AgNO_(3)` which SHOWS that one MOLECULE of the complex gives two `Cl^(-)` ions in solution. THUS the formula of the complex is `[Co(NH_(3))_(5)Cl]Cl_(2)` i.e. 1:2 electrolyge.


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