`4.5 g` of aluminium (at mass `27 u`) is deposited at cathode from `Al^(3+)` solution by a certain quantity of electric charge. The volume of hydrogen gas produced at `STP` from `H^(+)` ions in solution by the same quantity of electric charge will be:

Options

A. (a) 44.8 L
B. (b) 22.4 L
C. (c) 11.2 L
D. (d) 5.6 L

Explanation

To solve the problem of determining the volume of hydrogen gas produced at STP from the electrolysis of \( H^+ \) ions when 4.5 g of aluminum is deposited, we can follow these steps:

### Step 1: Calculate the equivalent weight of aluminum
The equivalent weight of a substance can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Atomic weight}}{n} \]
where \( n \) is the number of electrons transferred per atom during the reaction.

For aluminum (\( Al^{3+} \)):
- Atomic weight of aluminum = 27 g/mol
- \( n \) (number of electrons gained) = 3

Thus, the equivalent weight of aluminum is:
\[ \text{Equivalent weight of Al} = \frac{27}{3} = 9 \text{ g/equiv} \]

### Step 2: Calculate the mass of hydrogen produced using Faraday's second law
According to Faraday's second law of electrolysis:
\[ \frac{\text{Mass of Al deposited}}{\text{Mass of H produced}} = \frac{\text{Equivalent weight of Al}}{\text{Equivalent weight of H}} \]

Let \( m_H \) be the mass of hydrogen produced. The equivalent weight of hydrogen (\( H_2 \)) is calculated as follows:
- Atomic weight of \( H_2 \) = 2 g/mol
- \( n \) (number of electrons gained) = 2 (since \( 2H^+ + 2e^- \rightarrow H_2 \))

Thus, the equivalent weight of hydrogen is:
\[ \text{Equivalent weight of H}_2 = \frac{2}{2} = 1 \text{ g/equiv} \]

Now substituting the values into Faraday's equation:
\[ \frac{4.5 \text{ g}}{m_H} = \frac{9}{1} \]
Cross-multiplying gives:
\[ 4.5 = 9 \cdot m_H \]
Solving for \( m_H \):
\[ m_H = \frac{4.5}{9} = 0.5 \text{ g} \]

### Step 3: Calculate the volume of hydrogen gas at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Since 2 g of \( H_2 \) is 1 mole, we can find the volume of 0.5 g of \( H_2 \) as follows:
\[ \text{Volume of } H_2 = \frac{22.4 \text{ L}}{2 \text{ g}} \times 0.5 \text{ g} = 5.6 \text{ L} \]

### Final Answer
The volume of hydrogen gas produced at STP is:
\[ \text{Volume of } H_2 = 5.6 \text{ L} \]

Thus, the correct answer is option D: 5.6 L.