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                                    0.5 F of electric current was passed through 5 molar solution of AgNO_(3), CuSO_(4) " and " AlCl_(3) connected in series. Find out the concentration of each of the electrolyte after the electrolysis. | 
                            
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Answer» SOLUTION :Given : Quantity of electricity, Q = 0.5 F <BR> Concentration of solution, C = 5 M  Solution : (i) For `AgNO_(3)` 1 mol of `Ag^(+)=1F` `""0.5F= 0.5` mol of `AgNO_(3)` Concentration of `AgNO_(3)` after electrolysis `""=5-0.5` `""=4.5 M` (ii) For `CuSO_(4)` 1 mol of `CuSO_(4) " (or) " Cu^(2+)=2F` 2F = 1 M `CuSO_(4)` `0.5F=1/2 times 0.5=0.25M` Concentration of `CuSO_(4)` after electrolysis `""=5-0.25` `""=4.75M` (iii) For `AlCl_(3)` 1 mol of `AlCl_(3)" Br "Al^(3+)=3F` `3F=1M" "AlCl_(3)` `0.5=1/3 times 0.5` `""=0.5/3=0.167M` Concentration of `AlCl_(3)` after electrolysis `""=5-0.167` Concentration of `AlCl_(3)` after electrolysis `""=4.833M`  | 
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