At 298 K, the mole percentage of N_{2(g)} in | vedclass

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(D) According to Henry's Law, $P_{N_2} = K_H \cdot X_{N_2}$.First, calculate the partial pressure of $N_2$ in air: $P_{N_2} = 	ext{mole fraction of }

Vedclass · Accepted Answer

$9.35 	imes 10^{-5}$

Vedclass · Answer

(D) According to Henry's Law, $P_{N_2} = K_H \cdot X_{N_2}$.First, calculate the partial pressure of $N_2$ in air: $P_{N_2} = 	ext{mole fraction of } N_2 	imes 	ext{Total pressure} = 0.8 	imes 10 \ atm = 8 \ atm$.Convert the partial pressure to $mm \ Hg$: $P_{N_2} = 8 \ atm 	imes 760 \ mm \ Hg/atm = 6080 \ mm \ Hg$.Now, use Henry's Law to find the mole fraction $X_{N_2}$ in water: $X_{N_2} = \frac{P_{N_2}}{K_H} = \frac{6080}{6.5 	imes 10^7}$.$X_{N_2} = 9.35 	imes 10^{-5}$.

At $298 \ K$, the mole percentage of $N_{2(g)}$ in air is $80\%$. Water is in equilibrium with air at a pressure of $10 \ atm$. What is the mole fraction of $N_{2(g)}$ in water at $298 \ K$? ($K_H$ for $N_2$ is $6.5 \times 10^7 \ mm \ Hg$)

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