A vessel contains `O_(2)andH_(2)` in 2 : 1 molar ratio at 10 atm pressure then calculate ratio of their rate of diffusion.

Options

A. 1:4
B. 1:8
C. 1:2
D. 1:6

Explanation

To solve the problem of finding the ratio of the rates of diffusion of \( O_2 \) and \( H_2 \) in a vessel containing them in a 2:1 molar ratio at 10 atm pressure, we can follow these steps:

### Step 1: Understand the relationship between diffusion rates and molecular weights
According to Graham's law of effusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. The formula can be expressed as:

\[ \frac{\text{Rate of } O_2}{\text{Rate of } H_2} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \]

Where \( M_{H_2} \) is the molar mass of hydrogen and \( M_{O_2} \) is the molar mass of oxygen.

### Step 2: Determine the molar masses
The molar masses of the gases are:
- Molar mass of \( H_2 \) = 2 g/mol
- Molar mass of \( O_2 \) = 32 g/mol

### Step 3: Substitute the values into the formula
Now substituting the molar masses into the equation:

\[ \frac{\text{Rate of } O_2}{\text{Rate of } H_2} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

### Step 4: Calculate the ratio of the rates of diffusion
This means:

\[ \frac{\text{Rate of } O_2}{\text{Rate of } H_2} = \frac{1}{4} \]

To express this as a ratio, we can write:

\[ \text{Rate of } O_2 : \text{Rate of } H_2 = 1 : 4 \]

### Step 5: Finalize the answer
Thus, the ratio of the rates of diffusion of \( O_2 \) to \( H_2 \) is:

\[ \text{Rate of } O_2 : \text{Rate of } H_2 = 1 : 4 \]