Henry's law constant for the solubility of N_ | vedclass

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(A) Total pressure $P_T = 5\, atm$. Mole fraction of $N_2$ in air,$X_{N_2} = 0.6$. Partial pressure of $N_2$,$P_{N_2} = P_T 	imes X_{N_2} = 5 	imes

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$3.0 	imes 10^{-4}$

Vedclass · Answer

(A) Total pressure $P_T = 5\, atm$. Mole fraction of $N_2$ in air,$X_{N_2} = 0.6$. Partial pressure of $N_2$,$P_{N_2} = P_T 	imes X_{N_2} = 5 	imes 0.6 = 3\, atm$. According to Henry's law,$P_{N_2} = K_H 	imes X_{solute}$,where $X_{solute}$ is the mole fraction of $N_2$ in water. $X_{solute} = \frac{P_{N_2}}{K_H} = \frac{3\, atm}{1.0 	imes 10^5\, atm} = 3 	imes 10^{-5}$. Since the amount of $N_2$ is very small,$X_{solute} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}} \approx \frac{n_{N_2}}{n_{H_2O}}$. Given $n_{H_2O} = 10\, mol$,so $n_{N_2} = X_{solute} 	imes 10 = 3 	imes 10^{-5} 	imes 10 = 3.0 	imes 10^{-4}\, mol$.

Henry's law constant for the solubility of $N_2$ gas in water at $298\, K$ is $1.0 \times 10^5\, atm$. The mole fraction of $N_2$ in air is $0.6$. The moles of $N_2$ from air dissolved in $10$ moles of water at $298\, K$ and $5\, atm$ pressure is

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