At T(K),the partial pressure of dissolved oxy | vedclass

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(B) According to Henry's Law,$P_{O_2} = K_{H} \chi_{O_2}$.Given: $P_{O_2} = 1 \ bar$,$K_{H} = 50 \ kbar = 50 	imes 10^3 \ bar$.$\chi_{O_2} = \frac{P_

Vedclass · Accepted Answer

$35.50$

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(B) According to Henry's Law,$P_{O_2} = K_{H} \chi_{O_2}$.Given: $P_{O_2} = 1 \ bar$,$K_{H} = 50 \ kbar = 50 	imes 10^3 \ bar$.$\chi_{O_2} = \frac{P_{O_2}}{K_{H}} = \frac{1}{50 	imes 10^3} = 2 	imes 10^{-5}$.Since $\chi_{O_2} = \frac{n_{O_2}}{n_{O_2} + n_{H_2O}} \approx \frac{n_{O_2}}{n_{H_2O}}$,and $n_{H_2O} = \frac{1000 \ g}{18 \ g/mol} \approx 55.5 \ mol$.$n_{O_2} = \chi_{O_2} 	imes 55.5 = 2 	imes 10^{-5} 	imes 55.5 = 1.11 	imes 10^{-3} \ mol$.Mass of $O_2 = 1.11 	imes 10^{-3} \ mol 	imes 32 \ g/mol = 0.03552 \ g$.Concentration in $ppm = \frac{	ext{mass of solute}}{	ext{mass of solution}} 	imes 10^6 = \frac{0.03552 \ g}{1000 \ g} 	imes 10^6 = 35.52 \ ppm \approx 35.50 \ ppm$.

At $T(K)$,the partial pressure of dissolved oxygen in $1 \ L$ water is $1 \ bar$. The concentration of oxygen in $ppm$ is ($K_{H}$ of $O_2$ at $T(K)$ is $50 \ kbar$)

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