Aluminium oxide may be electrolysed at `1000^(@)`C to furnish aluminium metal (Atomic mass = 27 amu, 1 Faraday = 96500 Coulomb). The cathode reaction is `Al^(3+) + 3e^(-) rightarrow Al`. To prepare 5.12 kg of aluminium metal by this method would require:

Options

A. `5.49xx10^(1)C` of electricity
B. `5.49xx10^(4)C` of electricity
C. `1.83xx10^(7)C` o electricity
D. `5.49xx10^(7)C` of electricity

Explanation

To solve the problem of how much electricity is required to prepare 5.12 kg of aluminum metal through the electrolysis of aluminum oxide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of aluminum (W) = 5.12 kg = 5120 g (since 1 kg = 1000 g)
- Atomic mass of aluminum (A) = 27 amu
- Valency factor (n) = 3 (from the reaction: Al³⁺ + 3e⁻ → Al)
- Faraday's constant (F) = 96500 Coulombs

2. **Calculate the Equivalent Weight (E):**
The equivalent weight (E) can be calculated using the formula:
\[ E = \frac{A}{n} \]
Substituting the known values:
\[ E = \frac{27 \text{ g}}{3} = 9 \text{ g/equiv} \]

3. **Use Faraday's First Law of Electrolysis:**
According to Faraday's first law:
\[ W = Z \cdot Q \]
where Z is the electrochemical equivalent. We can express Z in terms of E and F:
\[ Z = \frac{E}{F} \]

4. **Rearranging the Equation for Quantity of Electricity (Q):**
From the above, we can rearrange to find Q:
\[ Q = \frac{W \cdot F}{E} \]

5. **Substituting Values into the Equation:**
Now, substituting the values we have:
\[ Q = \frac{5120 \text{ g} \cdot 96500 \text{ C}}{9 \text{ g/equiv}} \]

6. **Calculating Q:**
\[ Q = \frac{5120 \cdot 96500}{9} \]
\[ Q = \frac{494080000}{9} \approx 54908888.89 \text{ C} \]
Rounding this gives:
\[ Q \approx 5.49 \times 10^7 \text{ C} \]

### Final Answer:
The quantity of electricity required to prepare 5.12 kg of aluminum metal is approximately **5.49 x 10^7 Coulombs**.