Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect answer is option 'B'. Can you explain this answer?

Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity

Open App

JEE Exam > JEE Questions > Electrolysis of dilute aqueous NaCl solution ...

Join the Discussion on the App

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H₂ gas at the cathode is (1 Faraday = 96500 C mol^-1)

a)
9.65 x 10⁴sec
b)
19.3 x 10⁴sec
c)
28.95 x 10⁴sec
d)
38.6 x 10⁴sec

Correct answer is option 'B'. Can you explain this answer?

FREE This question is part of

WBJEE Chemistry Test - 12

Attempt Test

Attempt this Test

Verified Answer

Electrolysis of dilute aqueous NaCl solution was carried out by passin...

According to Faraday's first law of electrolysis:
m = EIt/F
Where:
m = mass of substance
E = equivalent weight of substance
I = current in amperes
t = time required
F = 96,500 Coulombs
t = mF EI = (0.01x 2g) x 96500 Coulomb

View all questions of this test

View all answers Start learning for free

< Previous Question Next Question >

Explore Courses for JEE exam

Similar JEE Doubts

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)

View answer

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)

View answer

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)

View answer

Electrolysis of dilute aqueous NaCI solution was carried out by passing 10 mA current. The time required to liberate 0.01 mole of H2 gas at the cathode is (1 F = 96500 C mol-1)(2008,Only One Option Correct Typ e)

View answer

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced.This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200; 1 Faraday = 96,500 coulombs)If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is Correct answer is '446'. Can you explain this answer?

View answer

Top Courses for JEEView all

Physics for JEE Main & Advanced

Mock Tests for JEE Main and Advanced 2025

Mathematics (Maths) for JEE Main & Advanced

Chemistry for JEE Main & Advanced

Chapter-wise Tests for JEE Main & Advanced

Top Courses for JEE

View all

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect answer is option 'B'. Can you explain this answer?

Share with a Friend

Answer this doubt

Question Description
Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect 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 carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect 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 carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect answer is option 'B'. Can you explain this answer?.

Solutions for Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x104secb)19.3 x104secc)28.95 x104secd)38.6 x104secCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.

Explore Courses for JEE exam