Question:
A gaseous mixture contains $\text{H}_2$ and $\text{O}_2$. The pressure of the mixture is $1\text{ bar}$. The weight percentage of $\text{O}_2$ is 80. The ratio of partial pressures of $\text{H}_2$ and $\text{O}_2$ is:
A gaseous mixture contains $\text{H}_2$ and $\text{O}_2$. The pressure of the mixture is $1\text{ bar}$. The weight percentage of $\text{O}_2$ is 80. The ratio of partial pressures of $\text{H}_2$ and $\text{O}_2$ is:
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For gas mixtures, convert mass percentages into moles first. Partial pressure ratios are always equal to mole ratios.
Updated On: Jun 15, 2026
- $5$
- $4$
- $0.2$
- $0.25$
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The Correct Option is B
Solution and Explanation
Concept:
According to Dalton's law,
\[
\frac{P_{\mathrm{H_2}}}{P_{\mathrm{O_2}}}
=
\frac{n_{\mathrm{H_2}}}{n_{\mathrm{O_2}}}
\]
Thus, we first calculate the mole ratio.
Step 1: Assume 100 g of mixture \[ m_{\mathrm{O_2}}=80\text{ g} \] \[ m_{\mathrm{H_2}}=20\text{ g} \]
Step 2: Calculate moles For hydrogen, \[ n_{\mathrm{H_2}} = \frac{20}{2} = 10 \] For oxygen, \[ n_{\mathrm{O_2}} = \frac{80}{32} = 2.5 \]
Step 3: Find pressure ratio \[ \frac{P_{\mathrm{H_2}}}{P_{\mathrm{O_2}}} = \frac{10}{2.5} = 4 \] Hence, \[ \boxed{4} \]
Step 1: Assume 100 g of mixture \[ m_{\mathrm{O_2}}=80\text{ g} \] \[ m_{\mathrm{H_2}}=20\text{ g} \]
Step 2: Calculate moles For hydrogen, \[ n_{\mathrm{H_2}} = \frac{20}{2} = 10 \] For oxygen, \[ n_{\mathrm{O_2}} = \frac{80}{32} = 2.5 \]
Step 3: Find pressure ratio \[ \frac{P_{\mathrm{H_2}}}{P_{\mathrm{O_2}}} = \frac{10}{2.5} = 4 \] Hence, \[ \boxed{4} \]
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