The ratio of the masses of methane and ethane in a gas mixture is `4:5`. The rate of number of their molecules in the mixture is:

Options

A. `4:5`
B. `3:2`
C. `2:3`
D. `5:4`

Answer

To solve the problem, we need to find the ratio of the number of molecules of methane (CH₄) and ethane (C₂H₆) in a gas mixture where their mass ratio is given as 4:5.

### Step-by-Step Solution:

1. **Identify the Molecular Masses**:
- The molecular mass of methane (CH₄) is calculated as follows:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol
- Molecular mass of CH₄ = 12 + (4 × 1) = 16 g/mol
- The molecular mass of ethane (C₂H₆) is calculated as follows:
- Carbon (C) = 12 g/mol (2 atoms)
- Hydrogen (H) = 1 g/mol (6 atoms)
- Molecular mass of C₂H₆ = (2 × 12) + (6 × 1) = 24 + 6 = 30 g/mol

2. **Set Up the Mass Ratio**:
- The mass ratio of methane to ethane is given as 4:5. This means:
- Mass of methane = 4x g
- Mass of ethane = 5x g
- Here, x is a common factor.

3. **Calculate the Number of Moles**:
- Number of moles of methane (n₁) = Mass / Molar mass
- \( n_1 = \frac{4x}{16} = \frac{x}{4} \)
- Number of moles of ethane (n₂) = Mass / Molar mass
- \( n_2 = \frac{5x}{30} = \frac{x}{6} \)

4. **Find the Ratio of Moles**:
- The ratio of the number of moles of methane to ethane is:
- \( \frac{n_1}{n_2} = \frac{\frac{x}{4}}{\frac{x}{6}} \)
- Simplifying this gives:
- \( \frac{n_1}{n_2} = \frac{x}{4} \times \frac{6}{x} = \frac{6}{4} = \frac{3}{2} \)

5. **Conclusion**:
- The ratio of the number of molecules of methane to ethane in the mixture is 3:2.

### Final Answer:
The rate of the number of their molecules in the mixture is **3:2**.