5.5 text{ mg} of nitrogen gas dissolves in 18 | vedclass

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(B) Moles of $N_2 = \frac{5.5 	imes 10^{-3} 	ext{ g}}{28 	ext{ g/mol}} \approx 1.96 	imes 10^{-4} 	ext{ mol}$.Moles of $H_2O = \frac{180 	ext{ g

Vedclass · Accepted Answer

$1 	imes 10^{-4}$

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(B) Moles of $N_2 = \frac{5.5 	imes 10^{-3} 	ext{ g}}{28 	ext{ g/mol}} \approx 1.96 	imes 10^{-4} 	ext{ mol}$.Moles of $H_2O = \frac{180 	ext{ g}}{18 	ext{ g/mol}} = 10 	ext{ mol}$.Mole fraction of $N_2$ at $1 	ext{ atm}$ $(x_1)$ = $\frac{n_{N_2}}{n_{H_2O} + n_{N_2}} \approx \frac{1.96 	imes 10^{-4}}{10} = 1.96 	imes 10^{-5}$.According to Henry's law,$p = K_H \cdot x$,which implies $x \propto p$.At $5 	ext{ atm}$ pressure,the new mole fraction $(x_2)$ is $x_2 = 5 	imes x_1 = 5 	imes 1.96 	imes 10^{-5} = 9.8 	imes 10^{-5} \approx 1 	imes 10^{-4}$.

Gas	$K_H$ (kbar at $298 \ K$)
$CO_2$	$1.67$
$Ar$	$40.3$
$HCHO$	$1.83 \times 10^{-5}$
$CH_4$	$0.413$

$5.5 \text{ mg}$ of nitrogen gas dissolves in $180 \text{ g}$ of water at $273 \text{ K}$ and $1 \text{ atm}$ pressure due to nitrogen gas. The mole fraction of nitrogen in $180 \text{ g}$ of water at $5 \text{ atm}$ nitrogen pressure is approximately:

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