A gas X is diffused `(1)/(sqrt2)`time slower than `SO_(2)` at same temperature calculate V.D of gas X :-

Options

A. 64
B. 128
C. 32
D. 16

Answer

To solve the problem of finding the vapor density of gas X, which diffuses \( \frac{1}{\sqrt{2}} \) times slower than \( SO_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between diffusion rates and molecular mass**:
According to Graham's Law of Effusion, the rates of diffusion of two gases are inversely proportional to the square roots of their molar masses (molecular weights). This can be expressed as:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( R_1 \) and \( R_2 \) are the rates of diffusion of gas X and \( SO_2 \) respectively, and \( M_1 \) and \( M_2 \) are their respective molar masses.

2. **Assign known values**:
- Let \( R_2 \) (the rate of diffusion of \( SO_2 \)) be 1 (as a reference).
- Given that gas X diffuses \( \frac{1}{\sqrt{2}} \) times slower than \( SO_2 \), we can express the rate of gas X as:
\[ R_1 = \frac{R_2}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]
- The molar mass of \( SO_2 \) is calculated as:
\[ M_{SO_2} = 32 + 32 = 64 \, \text{g/mol} \]

3. **Set up the equation using Graham's Law**:
Plugging the values into Graham's Law gives:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_{SO_2}}{M_X}} \]
Substituting the known values:
\[ \frac{\frac{1}{\sqrt{2}}}{1} = \sqrt{\frac{64}{M_X}} \]

4. **Square both sides to eliminate the square root**:
\[ \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{64}{M_X} \]
This simplifies to:
\[ \frac{1}{2} = \frac{64}{M_X} \]

5. **Rearrange to find \( M_X \)**:
Cross-multiplying gives:
\[ M_X = 64 \times 2 = 128 \, \text{g/mol} \]

6. **Calculate the vapor density**:
The vapor density (VD) is defined as:
\[ VD = \frac{M_X}{2} \]
Substituting the value of \( M_X \):
\[ VD = \frac{128}{2} = 64 \]

### Final Answer:
The vapor density of gas X is \( 64 \, \text{g/L} \).