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 molar mass.

### Step-by-Step Solution:

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

2. **Using Graham's Law:**
- According to Graham's law, the relationship 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 effusion of hydrogen and the unknown gas, respectively, and \( M_1 \) and \( M_2 \) are their molar masses.

3. **Setting Up the Equation:**
- Let \( R_H \) be the rate of effusion of hydrogen and \( R_X \) be the rate of effusion of the unknown gas.
- Given that \( R_X = \frac{1}{8} R_H \), we can substitute this into the equation:
\[ \frac{R_H}{R_X} = 8 \]
- Thus, we can write:
\[ 8 = \sqrt{\frac{M_X}{M_H}} \]

4. **Squaring Both Sides:**
- Squaring both sides to eliminate the square root gives:
\[ 64 = \frac{M_X}{M_H} \]

5. **Finding the Molar Mass of Hydrogen:**
- The molar mass of hydrogen (\( M_H \)) is approximately 2 g/mol (since H2 has two hydrogen atoms, each with a mass of approximately 1 g/mol).

6. **Calculating the Molar Mass of the Unknown Gas:**
- Rearranging the equation gives:
\[ M_X = 64 \times M_H \]
- Substituting the value of \( M_H \):
\[ M_X = 64 \times 2 = 128 \text{ g/mol} \]

7. **Conclusion:**
- The molecular mass of the unknown gas is 128 g/mol.

### Final Answer:
The molecular mass of the unknown gas is **128 g/mol**.

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