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

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(A) According to Henry's Law,$p_{N_2} = K_H 	imes x_{N_2}$.First,calculate the partial pressure of $N_2$ in the air: $p_{N_2} = x_{N_2(air)} 	imes P

Vedclass · Accepted Answer

$4 	imes 10^{-4}$

Vedclass · Answer

(A) According to Henry's Law,$p_{N_2} = K_H 	imes x_{N_2}$.First,calculate the partial pressure of $N_2$ in the air: $p_{N_2} = x_{N_2(air)} 	imes P_{total} = 0.8 	imes 5 \ atm = 4 \ atm$.Now,calculate the mole fraction of $N_2$ in the water: $x_{N_2(water)} = \frac{p_{N_2}}{K_H} = \frac{4 \ atm}{1.0 	imes 10^5 \ atm} = 4 	imes 10^{-5}$.Since $x_{N_2(water)} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}}$,and $n_{N_2}$ is very small compared to $n_{H_2O} = 10 \ moles$,we can approximate $x_{N_2(water)} \approx \frac{n_{N_2}}{n_{H_2O}}$.Therefore,$n_{N_2} = x_{N_2(water)} 	imes n_{H_2O} = (4 	imes 10^{-5}) 	imes 10 = 4 	imes 10^{-4} \ moles$.

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 $0.8$. Calculate the number of moles of $N_2$ dissolved in $10 \ moles$ of water at $298 \ K$ and a total pressure of $5 \ atm$.

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