Solubility of N_2 gas in water at 25,^{circ}C | vedclass

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(B) Henry's law states that $P = K_H 	imes X$,where $P$ is the partial pressure,$K_H$ is Henry's law constant,and $X$ is the mole fraction of the gas

Vedclass · Accepted Answer

$8.02 	imes 10^4\, atm$

Vedclass · Answer

(B) Henry's law states that $P = K_H 	imes X$,where $P$ is the partial pressure,$K_H$ is Henry's law constant,and $X$ is the mole fraction of the gas.First,calculate the mole fraction of $N_2$ $(X_{N_2})$.Given molarity $M = 5.4 	imes 10^{-4}\, M$,which means $5.4 	imes 10^{-4}\, mol$ of $N_2$ in $1\, L$ of water.Number of moles of water in $1\, L$ $(1000\, g)$ is $n_{H_2O} = \frac{1000}{18} \approx 55.56\, mol$.Since $n_{N_2} \ll n_{H_2O}$,$X_{N_2} \approx \frac{n_{N_2}}{n_{H_2O}} = \frac{5.4 	imes 10^{-4}}{55.56} \approx 9.72 	imes 10^{-6}$.Now,calculate $K_H = \frac{P_{N_2}}{X_{N_2}} = \frac{0.78\, atm}{9.72 	imes 10^{-6}} \approx 8.02 	imes 10^4\, atm$.

Solubility of $N_2$ gas in water at $25\,^{\circ}C$ and $0.78\, atm$ is $5.4 \times 10^{-4}\, M$. Calculate Henry's law constant for $N_2$ in water at $25\,^{\circ}C$.

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