The pressure of a mixture of equal weight of two gases of mol wt. 4 and 40, is 1.1 atm. The partial pressure of the lighter gas in this mixture is

Options

A. 1 atm
B. 0.1 atm
C. 0.15 atm
D. 0.5 atm

Explanation

To solve the problem of finding the partial pressure of the lighter gas in a mixture of two gases with molecular weights of 4 and 40, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Molecular weight of gas A (lighter gas) = 4 g/mol
- Molecular weight of gas B (heavier gas) = 40 g/mol
- Total pressure of the gas mixture = 1.1 atm
- Equal weight of both gases is considered.

2. **Assume Equal Weights:**
- Let’s assume we have 40 grams of each gas (this is a convenient choice since it is equal and will simplify calculations).

3. **Calculate the Number of Moles for Each Gas:**
- For gas A (molecular weight = 4 g/mol):
\[ \text{Moles of gas A} = \frac{\text{mass}}{\text{molecular weight}} = \frac{40 \, \text{g}}{4 \, \text{g/mol}} = 10 \, \text{moles} \]
- For gas B (molecular weight = 40 g/mol):
\[ \text{Moles of gas B} = \frac{\text{mass}}{\text{molecular weight}} = \frac{40 \, \text{g}}{40 \, \text{g/mol}} = 1 \, \text{mole} \]

4. **Calculate Total Moles in the Mixture:**
- Total moles = Moles of gas A + Moles of gas B
\[ \text{Total moles} = 10 + 1 = 11 \, \text{moles} \]

5. **Calculate the Mole Fraction of the Lighter Gas (Gas A):**
- Mole fraction of gas A (lighter gas) = \(\frac{\text{Moles of gas A}}{\text{Total moles}}\)
\[ \text{Mole fraction of gas A} = \frac{10}{11} \]

6. **Calculate the Partial Pressure of the Lighter Gas:**
- Using Dalton's Law of Partial Pressures:
\[ \text{Partial pressure of gas A} = \text{Mole fraction of gas A} \times \text{Total pressure} \]
\[ \text{Partial pressure of gas A} = \frac{10}{11} \times 1.1 \, \text{atm} \]
\[ \text{Partial pressure of gas A} = 1 \, \text{atm} \]

7. **Final Answer:**
- The partial pressure of the lighter gas (gas A) in the mixture is **1 atm**.