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Electrolysis of dilute aqueous NaCl solution was carriedout by passing 10 milli ampere current. The time required toliberate 0.01 mol of H2 gas at the cathode is (1 Faraday =96500 C mol–1)              (2008S)
  • a)
    9.65 × 104 sec
  • b)
    19.3 × 104 sec
  • c)
    28.95 × 104 sec
  • d)
    38.6 × 104 sec
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Electrolysis of dilute aqueous NaCl solution was carriedout by passing...
Give : I = 10 milliamperes ; IF = 96500 C mol–1
t = ? ; Moles of H2 produces = 0.01 mol
From the law of electrolysis, we have
Equivalents of H2 produces = 
Substituting given values, we get
i.e. (b) is the correct answer
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Electrolysis of dilute aqueous NaCl solution was carriedout by passing 10 milli ampere current. The time required toliberate 0.01 mol of H2 gas at the cathode is (1 Faraday =96500 C mol–1) (2008S)a)9.65 × 104 secb)19.3 × 104 secc)28.95 × 104 secd)38.6 × 104 secCorrect answer is option 'B'. Can you explain this answer?
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Electrolysis of dilute aqueous NaCl solution was carriedout by passing 10 milli ampere current. The time required toliberate 0.01 mol of H2 gas at the cathode is (1 Faraday =96500 C mol–1) (2008S)a)9.65 × 104 secb)19.3 × 104 secc)28.95 × 104 secd)38.6 × 104 secCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Electrolysis of dilute aqueous NaCl solution was carriedout by passing 10 milli ampere current. The time required toliberate 0.01 mol of H2 gas at the cathode is (1 Faraday =96500 C mol–1) (2008S)a)9.65 × 104 secb)19.3 × 104 secc)28.95 × 104 secd)38.6 × 104 secCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Electrolysis of dilute aqueous NaCl solution was carriedout by passing 10 milli ampere current. The time required toliberate 0.01 mol of H2 gas at the cathode is (1 Faraday =96500 C mol–1) (2008S)a)9.65 × 104 secb)19.3 × 104 secc)28.95 × 104 secd)38.6 × 104 secCorrect answer is option 'B'. Can you explain this answer?.
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