Equal weights of ethane and hydrogen are mixed in an empty container at `25^(@)C`. The fraction of the total pressure exerted by hydrogen is

Options

A. `1 : 2`
B. `1 : 1`
C. `1 : 16`
D. `15 : 16`

Explanation

To solve the problem of finding the fraction of the total pressure exerted by hydrogen when equal weights of ethane (C2H6) and hydrogen (H2) are mixed in a container at 25°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Molar Masses:**
- Molar mass of Hydrogen (H2) = 2 g/mol
- Molar mass of Ethane (C2H6) = 30 g/mol

2. **Assume Equal Weights:**
- Let the mass of hydrogen (H2) be \( X \) grams.
- Since the weights are equal, the mass of ethane (C2H6) will also be \( X \) grams.

3. **Calculate Moles of Each Gas:**
- Moles of hydrogen (H2) = \( \frac{X}{2} \)
- Moles of ethane (C2H6) = \( \frac{X}{30} \)

4. **Calculate Total Moles:**
- Total moles = Moles of H2 + Moles of C2H6
- Total moles = \( \frac{X}{2} + \frac{X}{30} \)

5. **Find a Common Denominator:**
- The common denominator for 2 and 30 is 30.
- Convert the moles:
- Moles of H2 = \( \frac{15X}{30} \)
- Moles of C2H6 = \( \frac{X}{30} \)

6. **Combine Moles:**
- Total moles = \( \frac{15X}{30} + \frac{X}{30} = \frac{16X}{30} \)

7. **Calculate Mole Fraction of Hydrogen:**
- Mole fraction of hydrogen (H2) = \( \frac{\text{Moles of H2}}{\text{Total moles}} \)
- Mole fraction of H2 = \( \frac{\frac{15X}{30}}{\frac{16X}{30}} \)
- The \( X \) and \( 30 \) cancel out:
- Mole fraction of H2 = \( \frac{15}{16} \)

8. **Calculate Fraction of Total Pressure Exerted by Hydrogen:**
- According to Dalton's Law of Partial Pressures, the partial pressure of hydrogen is given by:
- \( P_{H2} = \text{Mole fraction of H2} \times P_T \)
- Thus, the fraction of total pressure exerted by hydrogen is simply the mole fraction of hydrogen:
- Fraction of total pressure by H2 = \( \frac{15}{16} \)

### Final Answer:
The fraction of the total pressure exerted by hydrogen is \( \frac{15}{16} \).