Henry's law constant of oxygen is $1.4 \times 10^{-3}$ mol. lit $^{-1} . \mathrm{atm}^{-1}$ at $298…

Henry's law constant of oxygen is $1.4 \times 10^{-3}$ mol. lit $^{-1} . \mathrm{atm}^{-1}$ at $298 \mathrm{~K}$. How much of oxygen is dissolved in $100 \mathrm{~mL}$ at $298 \mathrm{~K}$
when the partial pressure of oxygen is $0.5 \mathrm{~atm} ?$
  1. $1.4 \mathrm{~g}$
  2. $3.2 \mathrm{~g}$
  3. $22.4 \mathrm{mg}$
  4. $2.24 \mathrm{mg}$

Solution

According to Henry's law, $\mathrm{m}=\mathrm{k} \times \mathrm{p}$
given $\mathrm{K}_{\mathrm{H}}=1.4 \times 10^{-3}$
$\mathrm{p}_{\mathrm{O}_{2}}=0.5 \mathrm{or}$
$\mathrm{p}_{\mathrm{O}_{2}}=\mathrm{K}_{\mathrm{H}} \times \mathrm{x}_{\mathrm{O}_{2}}$
$\therefore \mathrm{x}_{\mathrm{O}_{2}}=\frac{0.5}{1.4 \times 10^{-3}}$
No. of moles; $\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$
$0.7 \times 10^{-4}=\frac{\mathrm{m}}{32}$
$\mathrm{m}=22.4 \times 10^{-4} \mathrm{~g}=2.24 \mathrm{mg}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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