Question:
N$_2$ exerts a partial pressure of 7.648 bar when dissolved in 1 litre of water at 298 K. What is the mole fraction of N$_2$ at same temperature (Henry's law constant $(K_H)$ for N$_2$ at 298 K = 76.4 kbar)
N$_2$ exerts a partial pressure of 7.648 bar when dissolved in 1 litre of water at 298 K. What is the mole fraction of N$_2$ at same temperature (Henry's law constant $(K_H)$ for N$_2$ at 298 K = 76.4 kbar)
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Always ensure the units of pressure for $p$ and $K_H$ are the same before applying Henry's law.
Updated On: Apr 24, 2026
- $10^{-5}$
- $10^{-3}$
- $10^{-4}$
- $10^{-6}$
- $10^{-2}$
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The Correct Option is C
Solution and Explanation
Step 1:
\[ p = K_H \cdot \chi \;\Rightarrow\; \chi = \frac{p}{K_H} \]
Step 2:
\[ K_H = 76.4 \,\text{kbar} = 76.4 \times 10^3 \,\text{bar} = 7.64 \times 10^4 \,\text{bar} \]
Step 3:
\[ \chi = \frac{7.648}{7.64 \times 10^4} = \frac{7.648}{7.64} \times 10^{-4} \]
Step 4:
\[ \chi \approx 1.001 \times 10^{-4} \approx 10^{-4} \] Final Answer:
\[ Option (C) \; 10^{-4} \]
\[ p = K_H \cdot \chi \;\Rightarrow\; \chi = \frac{p}{K_H} \]
Step 2:
\[ K_H = 76.4 \,\text{kbar} = 76.4 \times 10^3 \,\text{bar} = 7.64 \times 10^4 \,\text{bar} \]
Step 3:
\[ \chi = \frac{7.648}{7.64 \times 10^4} = \frac{7.648}{7.64} \times 10^{-4} \]
Step 4:
\[ \chi \approx 1.001 \times 10^{-4} \approx 10^{-4} \] Final Answer:
\[ Option (C) \; 10^{-4} \]
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