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 will use Faraday's second law of electrolysis, which relates the mass of a substance deposited during electrolysis to the amount of electric charge passed through the electrolyte.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of aluminum deposited (m) = 0.09 g
- Time (t) = 96.5 s
- Molar mass of aluminum (Al) = 27 g/mol
- Faraday's constant (F) = 96500 C/mol
- n-factor for Al in AlCl3 = 3 (since Al goes from Al^3+ to Al)

2. **Calculate the Equivalent Weight of Aluminum:**
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{27 \text{ g/mol}}{3} = 9 \text{ g/equiv} \]

3. **Use Faraday's Second Law:**
The formula for Faraday's second law is given by:
\[ m = \frac{E \cdot F \cdot I \cdot t}{1000} \]
where:
- m = mass of the substance deposited (in grams)
- E = equivalent weight (in g/equiv)
- F = Faraday's constant (in C/mol)
- I = current (in amperes)
- t = time (in seconds)

4. **Rearranging the Formula:**
We need to find the current (I), so we rearrange the formula:
\[ I = \frac{m \cdot 1000}{E \cdot F \cdot t} \]

5. **Substituting the Values:**
Substitute the known values into the equation:
\[ I = \frac{0.09 \text{ g} \cdot 1000}{9 \text{ g/equiv} \cdot 96500 \text{ C/mol} \cdot 96.5 \text{ s}} \]

6. **Calculating the Current:**
\[ I = \frac{90}{9 \cdot 96500 \cdot 96.5} \]
Simplifying:
\[ I = \frac{90}{9 \cdot 96500 \cdot 96.5} = \frac{10}{96500 \cdot 96.5} \]
Now, calculating the denominator:
\[ 96500 \cdot 96.5 = 9311750 \]
Therefore:
\[ I = \frac{10}{9311750} \approx 10 \text{ A} \]

7. **Final Answer:**
The value of \( X \) (current) is approximately **10 A**.