Same mass of `CH_(4) and H_(2)` is taken in container. The partial pressure caused by `H_(2)` is

Options

A. `(8)/(9)`
B. `(1)/(9)`
C. `(1)/(2)`
D. 1

Answer

To solve the problem of finding the partial pressure caused by \( H_2 \) when equal masses of \( CH_4 \) and \( H_2 \) are taken in a container, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the mass of each gas:**
Let's assume we have 16 grams of each gas. This assumption simplifies calculations since the molar mass of \( CH_4 \) is 16 g/mol and that of \( H_2 \) is 2 g/mol.

2. **Calculate the number of moles of each gas:**
- For \( CH_4 \):
\[ \text{Number of moles of } CH_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mole} \]
- For \( H_2 \):
\[ \text{Number of moles of } H_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{16 \text{ g}}{2 \text{ g/mol}} = 8 \text{ moles} \]

3. **Calculate the total number of moles in the container:**
\[ \text{Total number of moles} = \text{moles of } CH_4 + \text{moles of } H_2 = 1 + 8 = 9 \text{ moles} \]

4. **Calculate the mole fraction of \( H_2 \):**
\[ \text{Mole fraction of } H_2 = \frac{\text{moles of } H_2}{\text{total moles}} = \frac{8}{9} \]

5. **Calculate the partial pressure of \( H_2 \):**
The partial pressure of a gas in a mixture can be calculated using Dalton's Law of Partial Pressures:
\[ P_{H_2} = \text{Mole fraction of } H_2 \times \text{Total pressure} \]
Assuming the total pressure in the container is 1 atm:
\[ P_{H_2} = \frac{8}{9} \times 1 \text{ atm} = \frac{8}{9} \text{ atm} \]

6. **Conclusion:**
The partial pressure caused by \( H_2 \) is \( \frac{8}{9} \text{ atm} \).

### Final Answer:
The partial pressure caused by \( H_2 \) is \( \frac{8}{9} \text{ atm} \).

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