Question

At T(K), a gaseous mixture contains H\(_2\) and O\(_2\). The total pressure of the mixture is 2 bar. The partial pressure of H\(_2\) is 1.778 bar. What is the weight (w/w) percentage of H\(_2\) in the mixture?

• A. 66.67
• B. 33.33
• C. 80.00
• D. 20.00

Hint:

Use the ratio of partial pressures to infer the mole ratio in gas mixtures, then multiply by molar masses to compute weight percentages.

Answer 1

The Correct Option is B

Step 1: Use Dalton’s Law of Partial Pressures.
Given: \[ P_{\text{total}} = 2 \, \text{bar}, \quad P_{\text{H}_2} = 1.778 \, \text{bar} \] So, \[ P_{\text{O}_2} = 2 - 1.778 = 0.222 \, \text{bar} \] Step 2: Assume ideal gas behavior and equal volume, use molar ratio based on pressure.
\[ \text{Moles of H}_2 \propto 1.778, \quad \text{Moles of O}_2 \propto 0.222 \] Step 3: Convert to masses.
\[ \text{Mass of H}_2 = 1.778 \times 2 = 3.556 \, \text{g} \quad (\text{since molar mass of H}_2 = 2) \] \[ \text{Mass of O}_2 = 0.222 \times 32 = 7.104 \, \text{g} \quad (\text{since molar mass of O}_2 = 32) \] Step 4: Calculate w/w % of H\(_2\).
\[ \frac{3.556}{3.556 + 7.104} \times 100 = \frac{3.556}{10.66} \times 100 \approx 33.33\% \]