A gas diffuses four times as quickly as oxygen. The molecular weight of the gas is

Options

A. 2
B. 4
C. 8
D. 16

Answer

To solve the problem of finding the molecular weight of a gas that diffuses four times as quickly as oxygen, we will use Graham's Law of Effusion/Diffusion. Here’s a step-by-step solution:

### Step 1: Understand Graham's Law
Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight. Mathematically, this can be expressed as:

\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]

where:
- \( r_1 \) = rate of diffusion of the unknown gas
- \( r_2 \) = rate of diffusion of oxygen
- \( M_1 \) = molecular weight of the unknown gas
- \( M_2 \) = molecular weight of oxygen (approximately 32 g/mol)

### Step 2: Set Up the Equation
According to the problem, the gas diffuses four times as quickly as oxygen. This means:

\[ r_1 = 4 \cdot r_2 \]

Substituting this into Graham's Law gives:

\[ \frac{4 \cdot r_2}{r_2} = \sqrt{\frac{32}{M_1}} \]

This simplifies to:

\[ 4 = \sqrt{\frac{32}{M_1}} \]

### Step 3: Square Both Sides
To eliminate the square root, we square both sides of the equation:

\[ 16 = \frac{32}{M_1} \]

### Step 4: Solve for \( M_1 \)
Now, we can solve for \( M_1 \) (the molecular weight of the unknown gas):

\[ M_1 = \frac{32}{16} \]

\[ M_1 = 2 \text{ g/mol} \]

### Conclusion
The molecular weight of the gas is 2 g/mol.