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 solve the problem of finding 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. 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 molar mass. Mathematically, this 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
In this problem, we know that gas 2 diffuses twice as rapidly as gas 1. Therefore, we can express this relationship as:

\[ r_2 = 2r_1 \]

### Step 3: Substitute into Graham's Law
Substituting \( r_2 \) into Graham's law gives us:

\[ \frac{r_1}{2r_1} = \sqrt{\frac{M_2}{M_1}} \]

This simplifies to:

\[ \frac{1}{2} = \sqrt{\frac{M_2}{M_1}} \]

### Step 4: Square Both Sides
To eliminate the square root, we square both sides:

\[ \left(\frac{1}{2}\right)^2 = \frac{M_2}{M_1} \]

This simplifies to:

\[ \frac{1}{4} = \frac{M_2}{M_1} \]

### Step 5: Substitute Known Values
We know that the molar mass of gas 1 (\( M_1 \)) is 64. Therefore, we substitute this value into the equation:

\[ \frac{1}{4} = \frac{M_2}{64} \]

### Step 6: Solve for \( M_2 \)
To find \( M_2 \), we multiply both sides by 64:

\[ M_2 = 64 \times \frac{1}{4} \]

Calculating this gives:

\[ M_2 = 16 \]

### Conclusion
Thus, the molecular weight of the gas that diffuses twice as rapidly as the gas with a molecular weight of 64 is **16**.