At 760 torr pressure and $20^{\circ} \mathrm{C}$ temperature, 11 of water dissolves $0.04 \mathrm{gm}$ of…
At 760 torr pressure and $20^{\circ} \mathrm{C}$ temperature, 11 of water dissolves $0.04 \mathrm{gm}$ of pure oxygen or $0.02 \mathrm{gm}$ of pure nitrogen. Assuming that dry air is composed of $20 \%$ oxygen and $80 \%$ nitrogen (by volume), the masses (in $\mathrm{g} / \mathrm{L}$ ) of oxygen and nitrogen dissolved by $1 \mathrm{~L}$ of water at $20^{\circ} \mathrm{C}$ exposed to air at a total pressure of 760 torr are
respectively:
$0.008,0.016$
$0.016,0.008$
$0.16,0.08$
$0.04,0.02$
Solution
Mass of oxygen $=\frac{20}{100} \times 0.04=0.008 \mathrm{~g} / \mathrm{L}$
Mass of nitrogen
$=\frac{80}{100} \times 0.02=0.016 \mathrm{~g} / \mathrm{L}$
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