Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of `H_(2)` gas at the cathode is (1 Faraday=96500 C `"mol"^(-1)`)A. `9.65xx10^(4)`secB. `19.3xx10^(4)` secC. `28.95xx10^(4)` secD. `38.6xx10^(4)` sec.

Electrolysis of dilute aqueous NaCl solution was carried out by passin

1.	Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of `H_(2)` gas at the cathode is (1 Faraday=96500 C `"mol"^(-1)`)A. `9.65xx10^(4)`secB. `19.3xx10^(4)` secC. `28.95xx10^(4)` secD. `38.6xx10^(4)` sec.
Answer» Correct Answer - B `NaCl+aqrarrNa++Cl^(-)` `H_(2)OhArrH^(+)+OH^(-)` `H_(2)^(+)+e^(-)rarr+1//2H_(2)` `therefore` 0.5 mole of `H_(2)` is liberated by 1F= 96500C 0.01 mole of `H_(2)` will be liberated by charge `=(96 500)/(0.5)xx0.01=1930C` `Q=1xxt` or `t=(Q)/(I)=(1930)/(10xx10^(-3))A` `=193000"sec" =19.3xx10^(4)` sec.

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Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of `H_(2)` gas at the cathode is (1 Faraday=96500 C `"mol"^(-1)`)A. `9.65xx10^(4)`secB. `19.3xx10^(4)` secC. `28.95xx10^(4)` secD. `38.6xx10^(4)` sec.

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