A direct current deposits deposits 54 g of silver (Atomic mass = 108) during electrolysis. How much aluminimum (Atomic mass =27) would be deposited from aluminium chloride solution by the same amount of electricity ?

Options

A. 4.5g
B. 5.4g
C. 54g
D. 27g

Answer

To solve the problem of how much aluminum would be deposited from aluminum chloride solution by the same amount of electricity that deposits 54 g of silver, we can follow these steps:

### Step 1: Identify the given data
- Weight of silver deposited (Ag) = 54 g
- Atomic mass of silver (Ag) = 108 g/mol
- Atomic mass of aluminum (Al) = 27 g/mol

### Step 2: Calculate the equivalent weight of silver
The equivalent weight of an element can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Atomic mass}}{\text{Valency}} \]
For silver (Ag), the valency is 1 (as it forms Ag+ ions):
\[ \text{Equivalent weight of Ag} = \frac{108}{1} = 108 \text{ g/equiv} \]

### Step 3: Calculate the equivalent weight of aluminum
For aluminum (Al), the valency is 3 (as it forms Al³⁺ ions):
\[ \text{Equivalent weight of Al} = \frac{27}{3} = 9 \text{ g/equiv} \]

### Step 4: Set up the relationship between the weights deposited
Using the formula for the relationship between the weights deposited:
\[ \frac{\text{Weight of Ag}}{\text{Weight of Al}} = \frac{\text{Equivalent weight of Ag}}{\text{Equivalent weight of Al}} \]
Substituting the known values:
\[ \frac{54}{\text{Weight of Al}} = \frac{108}{9} \]

### Step 5: Solve for the weight of aluminum deposited
Cross-multiplying gives:
\[ 54 \times 9 = 108 \times \text{Weight of Al} \]
This simplifies to:
\[ 486 = 108 \times \text{Weight of Al} \]
Now, divide both sides by 108:
\[ \text{Weight of Al} = \frac{486}{108} = 4.5 \text{ g} \]

### Conclusion
The weight of aluminum deposited from aluminum chloride solution by the same amount of electricity is **4.5 g**.

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