A certain gas effuses through a small opening of a vessel at a rate which is exactly one-fifth the rate at which helium does the same. Thus, the molecular weight of the gas is :

Options

A. 100
B. 75
C. 50
D. 25

Explanation

To solve the problem, we will use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molecular mass.

### Step-by-Step Solution:

1. **Understand the relationship given by Graham's Law**:
\[ \frac{R_g}{R_{He}} = \sqrt{\frac{M_{He}}{M_g}} \]
where \( R_g \) is the rate of effusion of the gas, \( R_{He} \) is the rate of effusion of helium, \( M_{He} \) is the molecular mass of helium, and \( M_g \) is the molecular mass of the gas we want to find.

2. **Identify the information provided**:
We know from the problem that the rate of effusion of the gas is one-fifth that of helium:
\[ R_g = \frac{1}{5} R_{He} \]

3. **Substitute \( R_g \) in the Graham's Law equation**:
Replacing \( R_g \) in the equation:
\[ \frac{\frac{1}{5} R_{He}}{R_{He}} = \sqrt{\frac{M_{He}}{M_g}} \]
This simplifies to:
\[ \frac{1}{5} = \sqrt{\frac{M_{He}}{M_g}} \]

4. **Square both sides to eliminate the square root**:
\[ \left(\frac{1}{5}\right)^2 = \frac{M_{He}}{M_g} \]
This gives us:
\[ \frac{1}{25} = \frac{M_{He}}{M_g} \]

5. **Rearranging the equation to find \( M_g \)**:
\[ M_g = 25 \cdot M_{He} \]

6. **Substituting the molecular mass of helium**:
The molecular mass of helium \( M_{He} \) is approximately 4 g/mol. Thus:
\[ M_g = 25 \cdot 4 = 100 \text{ g/mol} \]

7. **Conclusion**:
The molecular weight of the gas is 100 g/mol.

### Final Answer:
The molecular weight of the gas is **100 g/mol**.