Q22 / 160 · TS EAMCET 2024 (09 May Shift 2) · Chemistry
Dry air contains $79 \% \mathrm{~N}_2$ and $21 \% \mathrm{O}_2$. At $\mathrm{T}(\mathrm{K})$, if Henry's law constants for $\mathrm{N}_2$ and $\mathrm{O}_2$ in water are $8.57 \times 10^4 \mathrm{~atm}$ and $4.56 \times 10^4 \mathrm{~atm}$, the ratio of mole fractions of $\mathrm{N}_2$ and $\mathrm{O}_2$ dissolved in water at 1 atm is
- A$4: 1$
- B$1: 4$
- Correct answerC$2: 1$
- D$1: 2$
Solution
Total pressure of air over water $=1 \mathrm{~atm}$. Partial pressure of $\mathrm{N}_2$ and $\mathrm{O}_2$ are
$\begin{aligned}
& \mathrm{P}_{\mathrm{N}_2}=\frac{1 \times 79}{100}=0.79 \mathrm{~atm} \\
& \mathrm{P}_{\mathrm{O}_2}=\frac{1 \times 21}{100}=0.21 \mathrm{~atm}
\end{aligned}$
Applying Henry's law
$\begin{aligned}
& \mathrm{P}_{\mathrm{N}_2}=\mathrm{K}_{\mathrm{H}} \cdot \mathrm{X} \\
& \mathrm{X}_{\mathrm{N}_2}=\frac{\mathrm{P}_{\mathrm{N}_2}}{\mathrm{~K}_{\mathrm{N}_2}}=\frac{0.79}{8.54 \times 10^4}=9.25 \times 10^{-6} \\
& \mathrm{X}_{\mathrm{O}_2}=\frac{\mathrm{P}_{\mathrm{O}_2}}{\mathrm{~K}_{\mathrm{O}_2}}=\frac{0.21}{4.56 \times 10^4}=4.60 \times 10^{-6}
\end{aligned}$
$\begin{aligned} & \text { Proportion of } \mathrm{N}_2 \text { and } \mathrm{O}_2 \\ & \Rightarrow 9.25 \times 10^{-6}: 4.60 \times 10^{-6} \\ & \Rightarrow 2: 1\end{aligned}$
Hence, the correct answer is option: 'C'.