The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25. Then mole % of gas B in the mixture would be

Options

A. `25%`
B. `50%`
C. `75%`
D. `10%`

Answer

To solve the problem, we need to find the mole percent of gas B in a mixture of gas A and gas B, given the vapor density of the mixture.

### Step-by-Step Solution:

1. **Understand the Relationship Between Vapor Density and Molecular Weight:**
The vapor density (VD) of a gas is related to its molecular weight (MW) by the formula:
\[ \text{Vapor Density} = \frac{\text{Molecular Weight}}{2} \]
Given that the vapor density of the mixture is 25, we can find the molecular weight of the mixture.

2. **Calculate the Molecular Weight of the Mixture:**
\[ \text{Molecular Weight of the Mixture} = \text{Vapor Density} \times 2 = 25 \times 2 = 50 \]

3. **Set Up the Equation for Average Molecular Weight:**
Let \( x_A \) be the mole percent of gas A and \( x_B \) be the mole percent of gas B. Since the total mole percent must equal 100%, we have:
\[ x_B = 100 - x_A \]
The average molecular weight of the mixture can be expressed as:
\[ \text{Molecular Weight of the Mixture} = \frac{x_A \cdot M_A + x_B \cdot M_B}{x_A + x_B} \]
Substituting \( M_A = 40 \) and \( M_B = 80 \):
\[ 50 = \frac{x_A \cdot 40 + (100 - x_A) \cdot 80}{100} \]

4. **Simplify the Equation:**
Multiply both sides by 100:
\[ 5000 = 40x_A + (100 - x_A) \cdot 80 \]
Expanding the right side:
\[ 5000 = 40x_A + 8000 - 80x_A \]
Combine like terms:
\[ 5000 = 8000 - 40x_A \]
Rearranging gives:
\[ 40x_A = 8000 - 5000 \]
\[ 40x_A = 3000 \]
\[ x_A = \frac{3000}{40} = 75 \]

5. **Calculate the Mole Percent of Gas B:**
Since \( x_B = 100 - x_A \):
\[ x_B = 100 - 75 = 25 \]

### Final Answer:
The mole percent of gas B in the mixture is **25%**.

---