The Henry's law constant for the solubility o | vedclass

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(B) The partial pressure of $N_2$ is given by $P_{N_2} = \chi_{N_2} 	imes P_{total} = 0.8 	imes 5 \, atm = 4 \, atm$.According to Henry's law,$P_{N_

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$4 	imes 10^{-4}$

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(B) The partial pressure of $N_2$ is given by $P_{N_2} = \chi_{N_2} 	imes P_{total} = 0.8 	imes 5 \, atm = 4 \, atm$.According to Henry's law,$P_{N_2} = K_H 	imes x_{N_2}$,where $x_{N_2}$ is the mole fraction of $N_2$ in the solution.$4 = 1 	imes 10^5 	imes x_{N_2} \Rightarrow x_{N_2} = 4 	imes 10^{-5}$.Since $x_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}} \approx \frac{n_{N_2}}{n_{H_2O}}$ (as $n_{N_2}$ is very small compared to $n_{H_2O}$),$4 	imes 10^{-5} = \frac{n_{N_2}}{10} \Rightarrow n_{N_2} = 4 	imes 10^{-4} \, moles$.

The Henry's law constant for the solubility of $N_2$ gas in water at $25 \, ^oC$ is $1 \times 10^5 \, atm$. The mole fraction of $N_2$ in air is $0.8$. Calculate the number of moles of $N_2$ from air that will dissolve in $10 \, moles$ of $H_2O$ at $25 \, ^oC$ and at a total pressure of $5 \, atm$.

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