A certain gas takes 8 times as long to effuse out as compared to hydrogen gas for same volume at constant temperature and pressure. Its molecular mass will be

Options

A. 16 u
B. 32 u
C. 64 u
D. 128 u

Answer

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 weight.

### Step-by-step Solution:

1. **Understanding the Problem**:
We know that a certain gas takes 8 times as long to effuse compared to hydrogen gas. This means that the rate of effusion of hydrogen is 8 times that of the certain gas.

2. **Define Variables**:
- Let \( R_1 \) be the rate of effusion of the certain gas.
- Let \( R_2 \) be the rate of effusion of hydrogen.
- The molecular weight of hydrogen (\( M_2 \)) is 2 g/mol (since hydrogen gas \( H_2 \) has a molecular weight of 2 atomic mass units).
- Let \( M_1 \) be the molecular weight of the certain gas (which we need to find).

3. **Using Graham's Law**:
According to Graham's law:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]

4. **Relating the Rates**:
Since it takes 8 times longer for the certain gas to effuse, we can express the rates as:
\[ R_2 = 8R_1 \]
Therefore, we can substitute \( R_2 \) into Graham's law:
\[ \frac{R_1}{8R_1} = \sqrt{\frac{M_2}{M_1}} \]
This simplifies to:
\[ \frac{1}{8} = \sqrt{\frac{M_2}{M_1}} \]

5. **Squaring Both Sides**:
Squaring both sides gives:
\[ \left(\frac{1}{8}\right)^2 = \frac{M_2}{M_1} \]
\[ \frac{1}{64} = \frac{M_2}{M_1} \]

6. **Substituting the Molecular Weight of Hydrogen**:
We know \( M_2 = 2 \) g/mol:
\[ \frac{1}{64} = \frac{2}{M_1} \]

7. **Cross Multiplying to Solve for \( M_1 \)**:
Cross multiplying gives:
\[ M_1 = 2 \times 64 = 128 \text{ g/mol} \]

8. **Conclusion**:
The molecular mass of the certain gas is 128 atomic mass units (u).

### Final Answer:
The molecular mass of the certain gas is **128 u**.