A gas diffuse `(1)/(5)` times as fast as hydrogen at same pressure. Its molecular weight is

Options

A. 50
B. 25
C. `25sqrt(2)`
D. `50sqrt(2)`

Answer

To find the molecular weight of the gas that diffuses at a rate of \( \frac{1}{5} \) times that of hydrogen, we can use Graham's law of diffusion. Here’s a step-by-step solution:

### Step 1: Understand Graham's Law of Diffusion
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 (in this case, hydrogen)
- \( R_2 \) = rate of diffusion of gas 2 (the unknown gas)
- \( M_1 \) = molar mass of gas 1 (hydrogen)
- \( M_2 \) = molar mass of gas 2 (the unknown gas)

### Step 2: Set Up the Equation
Given that the gas diffuses at \( \frac{1}{5} \) the rate of hydrogen, we can write:

\[ \frac{R_{gas}}{R_{H_2}} = \frac{1}{5} \]

Let’s denote:
- \( R_{H_2} \) = rate of diffusion of hydrogen
- \( R_{gas} \) = rate of diffusion of the gas

Substituting into Graham's law gives us:

\[ \frac{1/5}{1} = \sqrt{\frac{M_{gas}}{M_{H_2}}} \]

### Step 3: Substitute Known Values
The molar mass of hydrogen (\( H_2 \)) is approximately 2 g/mol. Therefore, we can substitute \( M_{H_2} = 2 \) g/mol into the equation:

\[ \frac{1}{5} = \sqrt{\frac{M_{gas}}{2}} \]

### Step 4: Square Both Sides
To eliminate the square root, we square both sides of the equation:

\[ \left(\frac{1}{5}\right)^2 = \frac{M_{gas}}{2} \]

This simplifies to:

\[ \frac{1}{25} = \frac{M_{gas}}{2} \]

### Step 5: Solve for Molar Mass of the Gas
Now, we can solve for \( M_{gas} \):

\[ M_{gas} = 2 \times \frac{1}{25} = \frac{2}{25} \]

To find \( M_{gas} \), multiply both sides by 25:

\[ M_{gas} = 2 \times 25 = 50 \text{ g/mol} \]

### Conclusion
The molecular weight of the gas is **50 g/mol**.

---