A certain gas, G takes four times as long to effuse out as `H_(2)`. What is its molecular mass?

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 the Problem**:
- We know that gas G takes four times as long to effuse as hydrogen (H₂). This means that the rate of effusion of gas G is slower than that of hydrogen.

2. **Setting Up the Relationship**:
- Let the rate of effusion of hydrogen be \( R_{H_2} \).
- Since gas G takes four times as long to effuse, its rate of effusion \( R_G \) can be expressed as:
\[ R_G = \frac{R_{H_2}}{4} \]

3. **Applying Graham's Law**:
- According to Graham's Law:
\[ \frac{R_{H_2}}{R_G} = \sqrt{\frac{M_G}{M_{H_2}}} \]
- Where \( M_G \) is the molar mass of gas G and \( M_{H_2} \) is the molar mass of hydrogen (which is 2 g/mol).

4. **Substituting the Rates**:
- Substitute \( R_G \) in the equation:
\[ \frac{R_{H_2}}{\frac{R_{H_2}}{4}} = \sqrt{\frac{M_G}{2}} \]
- This simplifies to:
\[ 4 = \sqrt{\frac{M_G}{2}} \]

5. **Squaring Both Sides**:
- Square both sides to eliminate the square root:
\[ 16 = \frac{M_G}{2} \]

6. **Solving for Molar Mass of Gas G**:
- Multiply both sides by 2:
\[ M_G = 16 \times 2 = 32 \text{ g/mol} \]

### Final Answer:
The molecular mass of gas G is **32 g/mol**.

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