`O_(2)` and He are taken in equal weights in a vessel. The pressure exerted by Helium in the mixture is

Options

A. 1/8 th of total pressure
B. l/9th of total pressure
C. 2/9 th of total pressure
D. 8/9th of total pressure

Answer

To find the pressure exerted by helium in a mixture with oxygen when both gases are taken in equal weights, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Weight of Each Gas**:
Let the weight of each gas (Oxygen and Helium) be \( W \).

2. **Determine the Molar Masses**:
- The molar mass of Helium (He) is \( 4 \, \text{g/mol} \).
- The molar mass of Oxygen (O\(_2\)) is \( 32 \, \text{g/mol} \) (since \( O \) has a molar mass of \( 16 \, \text{g/mol} \) and there are two oxygen atoms in a molecule of oxygen).

3. **Calculate the Number of Moles**:
- The number of moles of Oxygen:
\[ n_{O_2} = \frac{W}{32} \]
- The number of moles of Helium:
\[ n_{He} = \frac{W}{4} \]

4. **Calculate the Total Number of Moles**:
- Total moles in the mixture:
\[ n_{total} = n_{O_2} + n_{He} = \frac{W}{32} + \frac{W}{4} \]
- To add these fractions, find a common denominator (which is 32):
\[ n_{total} = \frac{W}{32} + \frac{8W}{32} = \frac{9W}{32} \]

5. **Calculate the Mole Fraction of Helium**:
- Mole fraction of Helium (\( X_{He} \)):
\[ X_{He} = \frac{n_{He}}{n_{total}} = \frac{\frac{W}{4}}{\frac{9W}{32}} = \frac{W}{4} \times \frac{32}{9W} = \frac{8}{9} \]

6. **Calculate the Partial Pressure of Helium**:
- According to Dalton's Law of Partial Pressures, the partial pressure of a gas is given by:
\[ P_{He} = P_{total} \times X_{He} \]
- Therefore, the pressure exerted by Helium is:
\[ P_{He} = P_{total} \times \frac{8}{9} \]

### Final Answer:
The pressure exerted by Helium in the mixture is \( \frac{8}{9} \) of the total pressure.

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