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 molar mass.

### Step-by-Step Solution:

1. **Understanding Graham's Law**: According to Graham's law, the relationship between the rates of effusion of two gases is given by:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( R_1 \) and \( R_2 \) are the rates of effusion of gas 1 and gas 2, and \( M_1 \) and \( M_2 \) are their molar masses.

2. **Identifying the Gases**: In this problem, let:
- Gas 1 be the unknown gas (let's denote its molar mass as \( M_g \)).
- Gas 2 be helium (with a molar mass \( M_{He} = 4 \, \text{g/mol} \)).

3. **Setting Up the Equation**: We know from the problem that the rate of effusion of the unknown gas is one-fifth that of helium:
\[ R_g = \frac{1}{5} R_{He} \]
Substituting this into Graham's law gives:
\[ \frac{R_g}{R_{He}} = \frac{1}{5} \]

4. **Applying Graham's Law**: Now substituting into Graham's law:
\[ \frac{1/5}{1} = \sqrt{\frac{M_{He}}{M_g}} \]
This simplifies to:
\[ \frac{1}{5} = \sqrt{\frac{4}{M_g}} \]

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

6. **Cross-Multiplying**: Cross-multiply to find \( M_g \):
\[ M_g = 4 \times 25 = 100 \, \text{g/mol} \]

7. **Final Answer**: The molecular weight of the gas is:
\[ \boxed{100 \, \text{g/mol}} \]