Question

At \( T \)(K), hydrogen and oxygen gases are mixed in the ratio of \( 1:2 \) by mass in a closed vessel of volume \( V \) litres. If the total pressure of the gaseous mixture is \( p \) atm, the partial pressure of oxygen (in atm) is:

• A. \( \frac{p}{9} \)
• B. \( 9p \)
• C. \( \frac{8p}{9} \)
• D. \( \frac{p}{6} \)

Hint:

For gas mixtures, use Dalton’s Law: \[ P_{\text{gas}} = X_{\text{gas}} \cdot P_{\text{total}} \] where \( X_{\text{gas}} \) is the mole fraction.

Answer 1

The Correct Option is C

Step 1: Determining the Number of Moles Given the mass ratio of hydrogen to oxygen is \( 1:2 \), let: - Mass of \( H_2 \) = \( 1 \) g - Mass of \( O_2 \) = \( 2 \) g The molar masses are: - \( M_{H_2} = 2 \) g/mol - \( M_{O_2} = 32 \) g/mol Thus, number of moles: \[ n_{H_2} = \frac{1}{2} = 0.5 \text{ moles} \] \[ n_{O_2} = \frac{2}{32} = 0.0625 \text{ moles} \] Step 2: Computing Mole Fraction Total moles: \[ n_{\text{total}} = 0.5 + 0.0625 = 0.5625 \] Mole fraction of \( O_2 \): \[ X_{O_2} = \frac{0.0625}{0.5625} = \frac{1}{9} \] Step 3: Finding Partial Pressure Using Dalton’s Law: \[ P_{O_2} = X_{O_2} \cdot P \] \[ = \frac{8p}{9} \] Conclusion Thus, the correct answer is: \[ \frac{8p}{9} \]