The Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1.0…

The Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^5 \mathrm{~atm}$. The mole fraction of $\mathrm{N}_2$ in air is $0.8$. The number of moles of $\mathrm{N}_2$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and $5 \mathrm{~atm}$ pressure is
  1. $4 \times 10^{-4}$
  2. $4.0 \times 10^{-5}$
  3. $5.0 \times 10^{-4}$
  4. $4.0 \times 10^{-6}$

Solution

$P_{\mathrm{N}_2}=K_{\mathrm{H}} \times$ mole-fraction $\left(\mathrm{N}_2\right)$ mole-fraction $ \mathrm{N}_2 \frac{1}{10^5} \times 0.8 \times 5=4 \times 10^{-5} \mathrm{~mol}^{-1} $ In 10 mole solubility in $4 \times 10^{-4}$

Asked in: JEE Advanced 2009 (Paper 1)

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