JEE Exam > JEE Questions > 0.066 g of a metal was deposited when a curre...
Start Learning for Free
0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-
- a)31.8
- b)108.0
- c)8.0
- d)68.0
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
0.066 g of a metal was deposited when a current of 2 ampere passed for...
Total charge passed in Coulombs = It = 2 X100 = 200 Coulombs
200 Coulombs of charge deposits 0.066 g of metal.
Equivalent mass of the metal is that mass of metal deposited by 1 Farad or 96500 Coulombs.
Equivalent mass of metal
200 Coulombs of charge deposits 0.066 g of metal.
Equivalent mass of the metal is that mass of metal deposited by 1 Farad or 96500 Coulombs.
Equivalent mass of metal
Most Upvoted Answer
0.066 g of a metal was deposited when a current of 2 ampere passed for...
To determine the equivalent mass of the metal, we need to consider the Faraday's laws of electrolysis. According to Faraday's laws, the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
In this case, we are given the current (2 ampere) and the time (100 seconds) for which the current is passed through the solution. We are also given the mass of the metal deposited (0.066 g).
The formula to calculate the amount of substance deposited is:
Amount of substance (in moles) = Current (in amperes) × Time (in seconds) / Faraday's constant
The Faraday's constant is the charge of one mole of electrons and is equal to 96500 C/mol.
Let's calculate the amount of substance deposited:
Amount of substance (in moles) = 2 A × 100 s / 96500 C/mol
= 0.00207 mol
To find the equivalent mass of the metal, we divide the mass of the metal deposited by the amount of substance deposited:
Equivalent mass = Mass of metal deposited / Amount of substance deposited
= 0.066 g / 0.00207 mol
= 31.88 g/mol
Since the equivalent mass is given in grams per mole, we can round it to the nearest whole number to get 32 g/mol.
Therefore, the correct answer is option 'A' (31.8 g/mol).
In this case, we are given the current (2 ampere) and the time (100 seconds) for which the current is passed through the solution. We are also given the mass of the metal deposited (0.066 g).
The formula to calculate the amount of substance deposited is:
Amount of substance (in moles) = Current (in amperes) × Time (in seconds) / Faraday's constant
The Faraday's constant is the charge of one mole of electrons and is equal to 96500 C/mol.
Let's calculate the amount of substance deposited:
Amount of substance (in moles) = 2 A × 100 s / 96500 C/mol
= 0.00207 mol
To find the equivalent mass of the metal, we divide the mass of the metal deposited by the amount of substance deposited:
Equivalent mass = Mass of metal deposited / Amount of substance deposited
= 0.066 g / 0.00207 mol
= 31.88 g/mol
Since the equivalent mass is given in grams per mole, we can round it to the nearest whole number to get 32 g/mol.
Therefore, the correct answer is option 'A' (31.8 g/mol).
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer?
Question Description
0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer?.
0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer?.
Solutions for 0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. 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 0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for 0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of 0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-a)31.8b)108.0c)8.0d)68.0Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup