Molecular weight of a gas that diffuses twice as rapidly as the gas with molecular weight 64 is

Options

A. 16
B. 8
C. 64
D. 6.4

Answer

To find the molecular weight of a gas that diffuses twice as rapidly as a gas with a molecular weight of 64, we can use Graham's law of effusion/diffusion. Here's the 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 molar mass. Mathematically, it can be expressed as:
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]
where:
- \( r_1 \) = rate of diffusion of gas 1
- \( r_2 \) = rate of diffusion of gas 2
- \( M_1 \) = molar mass of gas 1
- \( M_2 \) = molar mass of gas 2

### Step 2: Set Up the Problem
Let:
- Gas 1 be the gas with molecular weight \( M_1 = 64 \)
- Gas 2 be the gas whose molecular weight we want to find, denoted as \( M_2 \)
- According to the problem, gas 1 diffuses twice as rapidly as gas 2, so we can write:
\[ r_1 = 2 \cdot r_2 \]

### Step 3: Substitute into Graham's Law
Substituting \( r_1 \) into Graham's law gives:
\[ \frac{2 \cdot r_2}{r_2} = \sqrt{\frac{M_2}{64}} \]
This simplifies to:
\[ 2 = \sqrt{\frac{M_2}{64}} \]

### Step 4: Square Both Sides
To eliminate the square root, square both sides:
\[ 2^2 = \frac{M_2}{64} \]
\[ 4 = \frac{M_2}{64} \]

### Step 5: Solve for \( M_2 \)
Now, multiply both sides by 64 to solve for \( M_2 \):
\[ M_2 = 4 \cdot 64 \]
\[ M_2 = 256 \]

### Conclusion
The molecular weight of the gas that diffuses twice as rapidly as the gas with a molecular weight of 64 is **256**.

---