Calculate K_H when 2 mmol of O_2 dissolve in | vedclass

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(C) Henry's Law is given by $P = K_H 	imes \chi$,where $\chi$ is the mole fraction of the gas in the solution. First,calculate the moles of water: $n

Vedclass · Accepted Answer

$1 	imes 10^{-4} \ bar$

Vedclass · Answer

(C) Henry's Law is given by $P = K_H 	imes \chi$,where $\chi$ is the mole fraction of the gas in the solution. First,calculate the moles of water: $n_{H_2O} = \frac{540 \ g}{18 \ g/mol} = 30 \ mol$. Since $2 \ mmol = 2 	imes 10^{-3} \ mol$ is very small,the total moles $\approx 30 \ mol$. Mole fraction $\chi_{O_2} = \frac{n_{O_2}}{n_{H_2O}} = \frac{2 	imes 10^{-3}}{30} = \frac{2}{3} 	imes 10^{-4} \approx 0.667 	imes 10^{-4}$. Using $K_H = \frac{P_{O_2}}{\chi_{O_2}} = \frac{2 	imes 10^{-8}}{0.667 	imes 10^{-4}} = 3 	imes 10^{-4} \ bar$.

Calculate $K_H$ when $2 \ mmol$ of $O_2$ dissolve in $540 \ mL$ of water at $27 \ ^oC$ temperature. Given $P_{O_2} = 2 \times 10^{-8} \ bar$.

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