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How many grams of silver could be plated out on a serving tray by passing electrolysis of a solution of Ag (1) salt for 8 hours at a current of 9 ampere? What is the area of the tray, if thickness of the silver platting is 0.002cm? Density of silver is 10g/cm^3.( Atomic mass of Ag=107.8)?
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Solution:
Given data:
Time, t = 8 hours = 8 x 3600 s
Current, I = 9 A
Atomic mass of Ag = 107.8 g/mol
Density of Ag = 10 g/cm³
Thickness of silver plating, d = 0.002 cm
Step 1: Calculation of the amount of charge passed
The amount of charge passed, Q = I x t
Q = 9 A x 8 x 3600 s
Q = 2.592 x 10⁵ C
Step 2: Calculation of the amount of Ag deposited
From Faraday's law of electrolysis, the amount of Ag deposited is given by:
m = (Q x M) / (n x F)
where,
m = mass of Ag deposited (in grams)
Q = amount of charge passed (in coulombs)
M = atomic mass of Ag (in grams/mole)
n = number of electrons required for the reduction of one Ag+ ion (1 electron in this case)
F = Faraday constant (96,485 C/mol)
Substituting the values, we get:
m = (2.592 x 10⁵ C x 107.8 g/mol) / (1 x 96,485 C/mol)
m = 287.9 g
Therefore, 287.9 g of Ag can be plated out on the serving tray.
Step 3: Calculation of the area of the tray
The volume of the Ag deposited is given by:
V = m / ρ
where,
V = volume of Ag deposited (in cm³)
m = mass of Ag deposited (in grams)
ρ = density of Ag (in g/cm³)
Substituting the values, we get:
V = 287.9 g / 10 g/cm³
V = 28.79 cm³
The area of the tray can be calculated using the formula:
V = A x d
where,
A = area of the tray (in cm²)
d = thickness of the silver plating (in cm)
Substituting the values, we get:
28.79 cm³ = A x 0.002 cm
A = 14,395 cm²
Therefore, the area of the tray is 14,395 cm².
Conclusion:
Thus, 287.9 g of Ag can be plated out on the serving tray by passing electrolysis of a solution of Ag(1) salt for 8 hours at a current of 9 ampere. The area of the tray is 14,395 cm² if the thickness of the silver plating is 0.002 cm.
Given data:
Time, t = 8 hours = 8 x 3600 s
Current, I = 9 A
Atomic mass of Ag = 107.8 g/mol
Density of Ag = 10 g/cm³
Thickness of silver plating, d = 0.002 cm
Step 1: Calculation of the amount of charge passed
The amount of charge passed, Q = I x t
Q = 9 A x 8 x 3600 s
Q = 2.592 x 10⁵ C
Step 2: Calculation of the amount of Ag deposited
From Faraday's law of electrolysis, the amount of Ag deposited is given by:
m = (Q x M) / (n x F)
where,
m = mass of Ag deposited (in grams)
Q = amount of charge passed (in coulombs)
M = atomic mass of Ag (in grams/mole)
n = number of electrons required for the reduction of one Ag+ ion (1 electron in this case)
F = Faraday constant (96,485 C/mol)
Substituting the values, we get:
m = (2.592 x 10⁵ C x 107.8 g/mol) / (1 x 96,485 C/mol)
m = 287.9 g
Therefore, 287.9 g of Ag can be plated out on the serving tray.
Step 3: Calculation of the area of the tray
The volume of the Ag deposited is given by:
V = m / ρ
where,
V = volume of Ag deposited (in cm³)
m = mass of Ag deposited (in grams)
ρ = density of Ag (in g/cm³)
Substituting the values, we get:
V = 287.9 g / 10 g/cm³
V = 28.79 cm³
The area of the tray can be calculated using the formula:
V = A x d
where,
A = area of the tray (in cm²)
d = thickness of the silver plating (in cm)
Substituting the values, we get:
28.79 cm³ = A x 0.002 cm
A = 14,395 cm²
Therefore, the area of the tray is 14,395 cm².
Conclusion:
Thus, 287.9 g of Ag can be plated out on the serving tray by passing electrolysis of a solution of Ag(1) salt for 8 hours at a current of 9 ampere. The area of the tray is 14,395 cm² if the thickness of the silver plating is 0.002 cm.
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How many grams of silver could be plated out on a serving tray by passing electrolysis of a solution of Ag (1) salt for 8 hours at a current of 9 ampere? What is the area of the tray, if thickness of the silver platting is 0.002cm? Density of silver is 10g/cm^3.( Atomic mass of Ag=107.8)?
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How many grams of silver could be plated out on a serving tray by passing electrolysis of a solution of Ag (1) salt for 8 hours at a current of 9 ampere? What is the area of the tray, if thickness of the silver platting is 0.002cm? Density of silver is 10g/cm^3.( Atomic mass of Ag=107.8)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about How many grams of silver could be plated out on a serving tray by passing electrolysis of a solution of Ag (1) salt for 8 hours at a current of 9 ampere? What is the area of the tray, if thickness of the silver platting is 0.002cm? Density of silver is 10g/cm^3.( Atomic mass of Ag=107.8)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How many grams of silver could be plated out on a serving tray by passing electrolysis of a solution of Ag (1) salt for 8 hours at a current of 9 ampere? What is the area of the tray, if thickness of the silver platting is 0.002cm? Density of silver is 10g/cm^3.( Atomic mass of Ag=107.8)?.
How many grams of silver could be plated out on a serving tray by passing electrolysis of a solution of Ag (1) salt for 8 hours at a current of 9 ampere? What is the area of the tray, if thickness of the silver platting is 0.002cm? Density of silver is 10g/cm^3.( Atomic mass of Ag=107.8)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about How many grams of silver could be plated out on a serving tray by passing electrolysis of a solution of Ag (1) salt for 8 hours at a current of 9 ampere? What is the area of the tray, if thickness of the silver platting is 0.002cm? Density of silver is 10g/cm^3.( Atomic mass of Ag=107.8)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How many grams of silver could be plated out on a serving tray by passing electrolysis of a solution of Ag (1) salt for 8 hours at a current of 9 ampere? What is the area of the tray, if thickness of the silver platting is 0.002cm? Density of silver is 10g/cm^3.( Atomic mass of Ag=107.8)?.
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