If N_2 gas is bubbled through water at 293 K | vedclass

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(A) According to Henry's law,the partial pressure of a gas is related to its mole fraction in the solution: $P = K_H 	imes x$.Given $P = 0.987 \ bar$

Vedclass · Accepted Answer

$0.716$

Vedclass · Answer

(A) According to Henry's law,the partial pressure of a gas is related to its mole fraction in the solution: $P = K_H 	imes x$.Given $P = 0.987 \ bar$ and $K_H = 76.48 \ kbar = 76480 \ bar$.Calculating the mole fraction of $N_2$ $(x_{N_2})$:$x_{N_2} = \frac{P}{K_H} = \frac{0.987 \ bar}{76480 \ bar} = 1.29 	imes 10^{-5}$.In $1 \ L$ of water,the number of moles of water is $n_{H_2O} = \frac{1000 \ g}{18 \ g/mol} = 55.5 \ mol$.Since $x_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}}$,and $n_{N_2} $n_{N_2} = x_{N_2} 	imes 55.5 \ mol = 1.29 	imes 10^{-5} 	imes 55.5 \ mol = 7.16 	imes 10^{-4} \ mol$.Converting to millimoles: $7.16 	imes 10^{-4} \ mol 	imes 1000 \ mmol/mol = 0.716 \ mmol$.

If $N_2$ gas is bubbled through water at $293 \ K$,how many millimoles of $N_2$ gas would dissolve in $1 \ L$ of water? Assume that $N_2$ exerts a partial pressure of $0.987 \ bar$. The Henry's law constant for $N_2$ at $293 \ K$ is $76.48 \ kbar$.

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