When X amperes of current is passed through molten `AlCl_(3)` for 96.5 s. 0.09 g of aluminium is deposited. What is the value of X?

Options

A. 10A
B. 20 A
C. 30 A
D. 40 A

Answer

To solve the problem, we need to determine the value of X (the current in amperes) when a certain amount of aluminum is deposited through electrolysis of molten AlCl₃. We will use Faraday's laws of electrolysis to find this value.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of aluminum deposited (m) = 0.09 g
- Time (t) = 96.5 s
- Faraday's constant (F) = 96500 C/mol

2. **Calculate the Equivalent Weight of Aluminum:**
- The atomic weight of aluminum (Al) = 27 g/mol.
- The n-factor for aluminum in AlCl₃ is 3 because aluminum can lose 3 electrons (Al → Al³⁺ + 3e⁻).
- Equivalent weight (E) = Atomic weight / n-factor = 27 g/mol / 3 = 9 g/equiv.

3. **Use Faraday's Second Law of Electrolysis:**
- According to Faraday's second law, the amount of substance deposited is given by:
\[ \text{Mass deposited} = \frac{E \cdot I \cdot t}{F} \]
- Rearranging this gives us:
\[ I = \frac{m \cdot F}{E \cdot t} \]

4. **Substitute the Values:**
- Substitute m = 0.09 g, E = 9 g/equiv, t = 96.5 s, and F = 96500 C/mol into the equation:
\[ I = \frac{0.09 \, \text{g} \cdot 96500 \, \text{C/mol}}{9 \, \text{g/equiv} \cdot 96.5 \, \text{s}} \]

5. **Calculate the Current (I):**
- Calculate the numerator:
\[ 0.09 \cdot 96500 = 8685 \, \text{g·C/mol} \]
- Calculate the denominator:
\[ 9 \cdot 96.5 = 868.5 \, \text{g·s/equiv} \]
- Now divide the two results:
\[ I = \frac{8685}{868.5} \approx 10 \, \text{A} \]

6. **Conclusion:**
- Therefore, the value of X (the current) is **10 A**.