If nearly `10^(5)` coulomb liberate `1 g` equivalent of aluminium, then the amount of aluminium (equivalent weight `9`) deposited through electrolysis in `20` minutes by a current of `50` amp will be

Options

A. `0.6 g`
B. `0.09 g`
C. `5.4 g`
D. `10.8 g`

Answer

To solve the problem, we will use Faraday's laws of electrolysis. According to the first law of electrolysis, the mass of a substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Charge (Q) that liberates 1 g equivalent of aluminum = \(10^5\) coulombs
- Equivalent weight of aluminum (Z) = 9 g/equiv
- Current (I) = 50 A
- Time (t) = 20 minutes

2. **Convert Time to Seconds:**
\[ t = 20 \text{ minutes} \times 60 \text{ seconds/minute} = 1200 \text{ seconds} \]

3. **Calculate the Total Charge (Q) Passed:**
Using the formula \(Q = I \times t\):
\[ Q = 50 \text{ A} \times 1200 \text{ s} = 60000 \text{ coulombs} \]

4. **Relate the Mass of Aluminum Deposited to Charge:**
According to the first law of electrolysis:
\[ \frac{m_1}{m_2} = \frac{Z_1 \cdot Q_2}{Z_2 \cdot Q_1} \]
Here, \(m_1 = 1 \text{ g}\), \(Z_1 = 9 \text{ g/equiv}\), \(Q_1 = 10^5 \text{ coulombs}\), and \(Q_2 = 60000 \text{ coulombs}\).

5. **Substituting the Values:**
We need to find \(m_2\) (mass of aluminum deposited):
\[ \frac{1 \text{ g}}{m_2} = \frac{9 \cdot 60000}{10^5} \]

6. **Cross Multiply to Solve for \(m_2\):**
\[ m_2 = \frac{1 \cdot 10^5}{9 \cdot 60000} \]

7. **Calculate \(m_2\):**
\[ m_2 = \frac{100000}{540000} = 5.4 \text{ g} \]

### Final Answer:
The amount of aluminum deposited through electrolysis in 20 minutes by a current of 50 A is **5.4 grams**.