IIT-JEE 2008 Paper 2 Offline | Electrochemistry Question 28 | Chemistry

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-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$$^{-1}$$].

9.65 $$\times$$ 10$$^4$$ sec

19.3 $$\times$$ 10$$^4$$ sec

28.95 $$\times$$ 10$$^4$$ sec

38.6 $$\times$$ 10$$^4$$ sec

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