During electrolysis of water, the volume of oxygen liberated is `2.24 dm^(3).` The volume of hydrogen liberated, under same conditions will be

Options

A. `2.24dm^(3)`
B. `1.12 dm^(3)`
C. `4.48 dm^(3)`
D. `0.56dm^(3)`

Answer

To solve the problem of finding the volume of hydrogen gas liberated during the electrolysis of water when the volume of oxygen liberated is given as 2.24 dm³, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The electrolysis of water can be represented by the following chemical equation:
\[ 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \]
This indicates that for every 2 moles of water, 2 moles of hydrogen gas and 1 mole of oxygen gas are produced.

2. **Volume Relationship**: According to Avogadro's law, at the same temperature and pressure, the volume of gas is directly proportional to the number of moles of the gas. From the balanced equation, we can see that:
- 2 moles of \(H_2\) are produced for every 1 mole of \(O_2\).
- Therefore, the volume of \(H_2\) produced will be twice the volume of \(O_2\) produced.

3. **Given Volume of Oxygen**: We are given that the volume of oxygen gas (\(O_2\)) liberated is 2.24 dm³.

4. **Calculate Volume of Hydrogen**: Since the volume of hydrogen gas (\(H_2\)) is twice that of oxygen gas, we can calculate it as follows:
\[ \text{Volume of } H_2 = 2 \times \text{Volume of } O_2 = 2 \times 2.24 \, \text{dm}^3 \]

5. **Perform the Calculation**:
\[ \text{Volume of } H_2 = 2 \times 2.24 = 4.48 \, \text{dm}^3 \]

6. **Final Answer**: The volume of hydrogen liberated during the electrolysis of water is 4.48 dm³.

### Summary:
The volume of hydrogen gas liberated during the electrolysis of water when 2.24 dm³ of oxygen is produced is **4.48 dm³**.