Hydrogen diffuses 6 times faster than the gas X, the molecular weight of X is

Options

A. 6
B. 36
C. 24
D. 72

Explanation

To solve the problem of finding the molecular weight of gas X, given that hydrogen diffuses 6 times faster than gas X, we can use Graham's Law of Effusion/Diffusion. Here’s a step-by-step solution:

### Step 1: Understand Graham's Law
Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where:
- \( R_1 \) = rate of diffusion of gas 1
- \( R_2 \) = rate of diffusion of gas 2
- \( M_1 \) = molar mass of gas 1
- \( M_2 \) = molar mass of gas 2

### Step 2: Assign Values
In our case:
- Let gas 1 be hydrogen (H₂) with a molar mass \( M_1 = 2 \, \text{g/mol} \).
- Let gas 2 be gas X with an unknown molar mass \( M_2 \).
- The rate of diffusion of hydrogen \( R_1 \) is 6 times that of gas X \( R_2 \), so we can express this as:
\[ R_1 = 6R_2 \]

### Step 3: Substitute Values into Graham's Law
Substituting the known values into Graham's Law gives:
\[ \frac{6R_2}{R_2} = \sqrt{\frac{M_2}{2}} \]
This simplifies to:
\[ 6 = \sqrt{\frac{M_2}{2}} \]

### Step 4: Square Both Sides
To eliminate the square root, we square both sides:
\[ 6^2 = \frac{M_2}{2} \]
This results in:
\[ 36 = \frac{M_2}{2} \]

### Step 5: Solve for \( M_2 \)
Now, we can solve for \( M_2 \) by multiplying both sides by 2:
\[ M_2 = 36 \times 2 = 72 \, \text{g/mol} \]

### Conclusion
The molecular weight of gas X is \( 72 \, \text{g/mol} \).