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On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)
- a)35.45 g
- b)70.9 g
- c)3.545 g
- d)17.77 g
Correct answer is option 'C'. Can you explain this answer?
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On passing 0.1 faraday of electricity through fused sodium chloride, t...
On passing 0.1F of electricity in NaCl, the amt of chloride liberated = 35.45*0.1 = 3.545
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On passing 0.1 faraday of electricity through fused sodium chloride, t...
**Answer:**
To determine the amount of chloride liberated when 0.1 faraday of electricity is passed through fused sodium chloride, we can use Faraday's law of electrolysis.
According to Faraday's law of electrolysis, the amount of substance liberated or deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The mathematical expression for this law is:
M = Q / (nF)
Where:
M is the mass of the substance liberated or deposited,
Q is the quantity of electricity passed through the electrolyte,
n is the number of moles of electrons transferred during the reaction,
F is the Faraday constant.
In this case, we are passing 0.1 faraday of electricity through fused sodium chloride. Since sodium chloride contains one chloride ion per formula unit, the number of moles of electrons transferred (n) is equal to 1.
So, using the equation, we can calculate the mass of chloride liberated:
M = (0.1 F) / (1 mol e- / F) = 0.1 mol
The atomic mass of chlorine (Cl) is given as 35.45 g/mol. Therefore, the mass of chloride liberated is:
M = 0.1 mol * 35.45 g/mol = 3.545 g
Hence, the correct answer is option C - 3.545 g.
To determine the amount of chloride liberated when 0.1 faraday of electricity is passed through fused sodium chloride, we can use Faraday's law of electrolysis.
According to Faraday's law of electrolysis, the amount of substance liberated or deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The mathematical expression for this law is:
M = Q / (nF)
Where:
M is the mass of the substance liberated or deposited,
Q is the quantity of electricity passed through the electrolyte,
n is the number of moles of electrons transferred during the reaction,
F is the Faraday constant.
In this case, we are passing 0.1 faraday of electricity through fused sodium chloride. Since sodium chloride contains one chloride ion per formula unit, the number of moles of electrons transferred (n) is equal to 1.
So, using the equation, we can calculate the mass of chloride liberated:
M = (0.1 F) / (1 mol e- / F) = 0.1 mol
The atomic mass of chlorine (Cl) is given as 35.45 g/mol. Therefore, the mass of chloride liberated is:
M = 0.1 mol * 35.45 g/mol = 3.545 g
Hence, the correct answer is option C - 3.545 g.
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On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)a)35.45 gb)70.9 gc)3.545 gd)17.77 gCorrect answer is option 'C'. Can you explain this answer?
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On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)a)35.45 gb)70.9 gc)3.545 gd)17.77 gCorrect answer is option 'C'. 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 On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)a)35.45 gb)70.9 gc)3.545 gd)17.77 gCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)a)35.45 gb)70.9 gc)3.545 gd)17.77 gCorrect answer is option 'C'. Can you explain this answer?.
On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)a)35.45 gb)70.9 gc)3.545 gd)17.77 gCorrect answer is option 'C'. 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 On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)a)35.45 gb)70.9 gc)3.545 gd)17.77 gCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)a)35.45 gb)70.9 gc)3.545 gd)17.77 gCorrect answer is option 'C'. Can you explain this answer?.
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On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)a)35.45 gb)70.9 gc)3.545 gd)17.77 gCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)a)35.45 gb)70.9 gc)3.545 gd)17.77 gCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chloride liberated is (At. mass of Cl = 35.45)a)35.45 gb)70.9 gc)3.545 gd)17.77 gCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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