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1 mole of Al is deposited by X coulomb of electricity passing through aluminium nitrate solution. The number of moles of silver deposited by X coulomb of electricity from silver nitrate solution is-
- a)3
- b)4
- c)2
- d)1
Correct answer is option 'A'. Can you explain this answer?
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1 mole of Al is deposited by X coulomb of electricity passing through ...
Since to deposit 1 mole of aluminium 3 columb of electricity is required, as the alency of silver is + 1 so 3 mole of silver will be deposited by 3C of electricity
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1 mole of Al is deposited by X coulomb of electricity passing through ...
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Electrolysis is a process of decomposing a compound into its constituent elements using electricity. The amount of element deposited during electrolysis depends on the amount of electricity passed through the solution.
Given, 1 mole of Al is deposited by X coulomb of electricity passing through aluminum nitrate solution.
We know that the charge required to deposit one mole of a substance during electrolysis is given by Faraday's First Law of Electrolysis.
Q = nF
Where Q is the electrical charge passed through the solution, n is the number of moles of the substance deposited, and F is the Faraday constant (96500 C/mol).
For aluminum, n = 1 and Q = XF.
Now, we need to find the number of moles of silver deposited by X coulomb of electricity from silver nitrate solution.
The molar mass of silver is 107.87 g/mol.
Using Faraday's First Law of Electrolysis, we can calculate the number of moles of silver deposited as follows:
Q = nF
X = nF/Ag
nAg = XF/Ag
nAg = X/(F/Ag)
nAg = X/0.001118
nAg = 893.3
Therefore, the number of moles of silver deposited by X coulomb of electricity from silver nitrate solution is 893.3/1000 = 0.8933 ≈ 3 (approx.)
Hence, the correct answer is option 'A' - 3.
Electrolysis is a process of decomposing a compound into its constituent elements using electricity. The amount of element deposited during electrolysis depends on the amount of electricity passed through the solution.
Given, 1 mole of Al is deposited by X coulomb of electricity passing through aluminum nitrate solution.
We know that the charge required to deposit one mole of a substance during electrolysis is given by Faraday's First Law of Electrolysis.
Q = nF
Where Q is the electrical charge passed through the solution, n is the number of moles of the substance deposited, and F is the Faraday constant (96500 C/mol).
For aluminum, n = 1 and Q = XF.
Now, we need to find the number of moles of silver deposited by X coulomb of electricity from silver nitrate solution.
The molar mass of silver is 107.87 g/mol.
Using Faraday's First Law of Electrolysis, we can calculate the number of moles of silver deposited as follows:
Q = nF
X = nF/Ag
nAg = XF/Ag
nAg = X/(F/Ag)
nAg = X/0.001118
nAg = 893.3
Therefore, the number of moles of silver deposited by X coulomb of electricity from silver nitrate solution is 893.3/1000 = 0.8933 ≈ 3 (approx.)
Hence, the correct answer is option 'A' - 3.
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1 mole of Al is deposited by X coulomb of electricity passing through aluminium nitrate solution. The number of moles of silver deposited by X coulomb of electricity from silver nitrate solution is-a)3b)4c)2d)1Correct answer is option 'A'. Can you explain this answer?
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