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

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

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$4.0 	imes 10^{-4}$

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(B) According to Henry's Law,$P_{N_2} = K_H 	imes X_{N_2}$.First,calculate the partial pressure of $N_2$ in air: $P_{N_2} = 	ext{mole fraction of } N_2 	imes 	ext{Total pressure} = 0.8 	imes 5 \ atm = 4 \ atm$.Now,calculate the mole fraction of $N_2$ in water $(X_{N_2})$: $X_{N_2} = P_{N_2} / K_H = 4 \ atm / (1.0 	imes 10^5 \ atm) = 4 	imes 10^{-5}$.Since $X_{N_2} = n_{N_2} / (n_{N_2} + n_{H_2O})$ and $n_{N_2} \ll n_{H_2O}$,we can approximate $X_{N_2} \approx n_{N_2} / n_{H_2O}$.Given $n_{H_2O} = 10 \ moles$,we have $n_{N_2} = X_{N_2} 	imes n_{H_2O} = (4 	imes 10^{-5}) 	imes 10 = 4 	imes 10^{-4} \ moles$.

The 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 $0.8$. The number of moles of $N_2$ from air dissolved in $10 \ moles$ of water at $298 \ K$ and $5 \ atm$ pressure will be:

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