A flask coins hydrogen and oxygen in the ratio of `2:1` by mass at temperature `27^(@)C`.The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is :

Options

A. ` 2 : 1`
B. ` 1 : 4`
C. ` 1 : 1`
D. `4 : 1`

Answer

To solve the problem, we need to find the ratio of the average kinetic energy per molecule of hydrogen (H₂) and oxygen (O₂) at a given temperature.

### Step-by-Step Solution:

1. **Understanding Average Kinetic Energy**:
The average kinetic energy (KE) per molecule of a gas is given by the formula:
\[ KE = \frac{3}{2} k_B T \]
where \( k_B \) is the Boltzmann constant and \( T \) is the absolute temperature in Kelvin.

2. **Identify the Gases and Their Molar Masses**:
- For Hydrogen (H₂): The molar mass is approximately \( 2 \, \text{g/mol} \).
- For Oxygen (O₂): The molar mass is approximately \( 32 \, \text{g/mol} \).

3. **Given Mass Ratio**:
The problem states that the mass ratio of hydrogen to oxygen is \( 2:1 \).

4. **Temperature Conversion**:
The temperature is given as \( 27^\circ C \). We need to convert this to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

5. **Calculating Average Kinetic Energies**:
Since the temperature is the same for both gases, we can write the average kinetic energy for hydrogen and oxygen as:
\[ KE_{H_2} = \frac{3}{2} k_B T \]
\[ KE_{O_2} = \frac{3}{2} k_B T \]

6. **Finding the Ratio of Average Kinetic Energies**:
The ratio of average kinetic energy per molecule of hydrogen to that of oxygen is:
\[ \text{Ratio} = \frac{KE_{H_2}}{KE_{O_2}} = \frac{\frac{3}{2} k_B T}{\frac{3}{2} k_B T} = 1 \]

7. **Final Result**:
Therefore, the ratio of average kinetic energy per molecule of hydrogen to oxygen is:
\[ \text{Ratio} = 1:1 \]

### Conclusion:
The final answer is that the ratio of average kinetic energy per molecule of hydrogen and oxygen is \( 1:1 \).