Equal weight of `CH_(4) " and " H_(2)` are mixed in an empty container at `25^(@)C`. The fraction of the total pressure exerted by `H_(2)` is

Options

A. `1//9`
B. `1//2`
C. `8//9`
D. `16//17`

Explanation

To solve the problem of finding the fraction of the total pressure exerted by hydrogen (H₂) when equal weights of methane (CH₄) and hydrogen are mixed in a container at 25°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Equal weights of CH₄ and H₂ are mixed.
- Let the weight of each gas be \( W \).

2. **Calculate the Molar Masses:**
- Molar mass of H₂ = 2 g/mol (since it has 2 hydrogen atoms).
- Molar mass of CH₄ = 16 g/mol (1 carbon atom + 4 hydrogen atoms).

3. **Calculate the Number of Moles:**
- Number of moles of H₂:
\[ n_{H_2} = \frac{W}{2} \]
- Number of moles of CH₄:
\[ n_{CH_4} = \frac{W}{16} \]

4. **Calculate the Total Moles:**
- Total moles \( n_{total} \):
\[ n_{total} = n_{H_2} + n_{CH_4} = \frac{W}{2} + \frac{W}{16} \]
- To add these fractions, find a common denominator (16):
\[ n_{total} = \frac{8W}{16} + \frac{W}{16} = \frac{9W}{16} \]

5. **Calculate the Mole Fraction of H₂:**
- Mole fraction of H₂ \( (X_{H_2}) \):
\[ X_{H_2} = \frac{n_{H_2}}{n_{total}} = \frac{\frac{W}{2}}{\frac{9W}{16}} = \frac{W}{2} \times \frac{16}{9W} = \frac{16}{18} = \frac{8}{9} \]

6. **Calculate the Fraction of Total Pressure Exerted by H₂:**
- The fraction of total pressure exerted by H₂ is equal to its mole fraction:
\[ \text{Fraction of pressure by H₂} = X_{H_2} = \frac{8}{9} \]

### Final Answer:
The fraction of the total pressure exerted by H₂ is \( \frac{8}{9} \).

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