1 mole of Al is deposited by X coulomb of electricity passing through molten aluminium nitrate solution. The number of moles of silver deosited by X coulomb of electricity from silver nitrate solution is:

Options

A. 3 moles
B. 4 moles
C. 2 moles
D. 1moles

Answer

To solve the problem, we need to determine how many moles of silver (Ag) are deposited when the same amount of electricity (X coulombs) is passed through a silver nitrate solution, given that 1 mole of aluminum (Al) is deposited from aluminum nitrate solution using X coulombs.

### Step-by-Step Solution:

1. **Determine the number of electrons required for aluminum deposition:**
- Aluminum ion (Al³⁺) requires 3 electrons to be reduced to aluminum metal (Al).
- Therefore, to deposit 1 mole of aluminum, we need 3 moles of electrons.

2. **Relate charge (coulombs) to moles of electrons:**
- According to Faraday's laws of electrolysis, the amount of charge (Q) required to deposit a certain amount of substance is given by:
\[ Q = n \cdot F \]
where \( n \) is the number of moles of electrons and \( F \) is Faraday's constant (approximately 96500 C/mol).
- For 1 mole of Al, the charge required is:
\[ Q = 3 \cdot F \]
- Since we are given that this charge is X coulombs, we can equate:
\[ X = 3 \cdot F \]

3. **Determine the number of moles of silver deposited:**
- For silver (Ag), the silver ion (Ag⁺) requires 1 electron to be reduced to silver metal (Ag).
- Therefore, to deposit 1 mole of silver, we need 1 mole of electrons.
- The number of moles of silver deposited (nₐg) can be calculated using the same charge (X) that was used for aluminum:
\[ n_{Ag} = \frac{X}{F} \]
- Since we know that \( X = 3 \cdot F \) from step 2, we can substitute this into the equation for silver:
\[ n_{Ag} = \frac{3 \cdot F}{F} = 3 \]

4. **Conclusion:**
- Therefore, the number of moles of silver deposited by X coulombs of electricity is:
\[ n_{Ag} = 3 \text{ moles} \]

### Final Answer:
The number of moles of silver deposited by X coulombs of electricity is **3 moles**.