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

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(A) Total pressure $P_T = 5 \ atm$. The mole fraction of nitrogen in air is $0.8$. Partial pressure of nitrogen $P_{N_2} = P_T 	imes X_{air} = 5 	im

Vedclass · Accepted Answer

$4.0 	imes 10^{-4}$

Vedclass · Answer

(A) Total pressure $P_T = 5 \ atm$. The mole fraction of nitrogen in air is $0.8$. Partial pressure of nitrogen $P_{N_2} = P_T 	imes X_{air} = 5 	imes 0.8 = 4 \ atm$. According to Henry's law,$P_{N_2} = K_H \cdot X_{sol}$,where $X_{sol}$ is the mole fraction of $N_2$ in the solution. $X_{sol} = \frac{P_{N_2}}{K_H} = \frac{4}{1.0 	imes 10^5} = 4 	imes 10^{-5}$. Since $X_{sol} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}} \approx \frac{n_{N_2}}{n_{H_2O}}$ because $n_{N_2} \ll n_{H_2O}$. Given $n_{H_2O} = 10 \ moles$,we have $4 	imes 10^{-5} = \frac{n_{N_2}}{10}$. Therefore,$n_{N_2} = 4 	imes 10^{-4} \ moles$.

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 is

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