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Q.
How much oxygen is dissolved in $100\, mL$ water at $298\, K$ if partial pressure of oxygen is $0.5\, atm$ and $K_H = 1.4 \times 10^{-3}\, mol/L/atm$ ?
Solutions
Solution:
According to Henry’s law, $s = K_H \times p$, where s is concentration of $O_2$ dissolved.
$s = 1.4 \times 10^{-3} \times 0.5$
$= 0.7 \times 10^{-3}\,mol/L$
$s = \frac{n}{V}$ or $n = 0.7 \times 10^{-3} \times 0.1$
$= 0.7 \times 10^{-4}\,mol$
$n = \frac{w}{M}$ or $w = n \times M$
$= 0.7 \times 10^{-4} \times 32$
$ = 22.4 \times 10^{-4}\, g$ or $2.24\, mg$