Question

At T(K), a gaseous mixture contains H\(_2\) and O\(_2\). The total pressure of the mixture is 2 bar. The weight percentage (w/w) of H\(_2\) is 33.33%. What is the approximate ratio of partial pressure of H\(_2\) and O\(_2\)?

• A. 8 : 1
• B. 4 : 1
• C. 2 : 1
• D. 3 : 1

Hint:

When a gas mixture's composition is given in weight percent, convert it to moles before applying Dalton’s Law of Partial Pressures. This allows accurate calculation of partial pressures.

Answer 1

The Correct Option is A

Step 1: Assume 100 g of mixture.
Given that H\(_2\) is 33.33% by weight, then: \[ \text{Mass of H}_2 = 33.33\ \text{g}, \quad \text{Mass of O}_2 = 66.67\ \text{g} \] Step 2: Convert masses to moles.
\[ \text{Moles of H}_2 = \frac{33.33}{2} = 16.665
\text{Moles of O}_2 = \frac{66.67}{32} \approx 2.083 \] Step 3: Use mole ratio to find partial pressure ratio.
Total moles = \(16.665 + 2.083 \approx 18.748\)
\[ \text{Partial pressure of H}_2 = \frac{16.665}{18.748} \times 2 \approx 1.777 \text{ bar}
\text{Partial pressure of O}_2 = \frac{2.083}{18.748} \times 2 \approx 0.223 \text{ bar} \] Step 4: Calculate the pressure ratio.
\[ \frac{P_{\text{H}_2}}{P_{\text{O}_2}} = \frac{1.777}{0.223} \approx 8 : 1 \]