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 solve the problem of finding the molecular weight of a gas that diffuses at a rate of \( \frac{1}{5} \) times that of hydrogen at the same pressure, we will use Graham's law of diffusion.

### Step-by-Step Solution:

1. **Understanding 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 (molecular weight). Mathematically, 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 1 and gas 2, and \( M_1 \) and \( M_2 \) are their respective molar masses.

2. **Assigning Values**:
Let:
- \( R_1 \) = rate of diffusion of the unknown gas
- \( R_2 \) = rate of diffusion of hydrogen
- \( M_1 \) = molecular weight of hydrogen = 2 g/mol
- \( M_2 \) = molecular weight of the unknown gas (which we need to find)

According to the problem, we have:
\[ R_1 = \frac{1}{5} R_2 \]

3. **Setting Up the Equation**:
Plugging the values into Graham's law:
\[ \frac{R_1}{R_2} = \frac{1}{5} = \sqrt{\frac{M_2}{2}} \]

4. **Squaring Both Sides**:
To eliminate the square root, we square both sides of the equation:
\[ \left(\frac{1}{5}\right)^2 = \frac{M_2}{2} \]
This simplifies to:
\[ \frac{1}{25} = \frac{M_2}{2} \]

5. **Cross-Multiplying**:
Cross-multiplying gives:
\[ M_2 = 2 \times \frac{1}{25} = \frac{2}{25} \]

6. **Calculating Molar Mass**:
To find \( M_2 \), we multiply both sides by 25:
\[ M_2 = 2 \times 25 = 50 \text{ g/mol} \]

Thus, the molecular weight of the unknown gas is **50 g/mol**.

### Final Answer:
The molecular weight of the gas is **50 g/mol**.