At $15^{\circ} \mathrm{C}$ and 1 atmosphere partial pressure of hydrogen, $20 \mathrm{~mL}$ of hydrogen…
At $15^{\circ} \mathrm{C}$ and 1 atmosphere partial pressure of hydrogen, $20 \mathrm{~mL}$ of hydrogen measured at STP dissolves in $1 \mathrm{~L}$ of water. If water at $15^{\circ} \mathrm{C}$ is exposed to a gaseous mixture having a total pressure of $1500 \mathrm{~mm}$ of $\mathrm{Hg}$ (excluding the vapour pressure of water) and containing $80 \%$ hydrogen by volumne, then the volume of hydrogen measured at STP that will dissolve in 1 L of water is
$20.0 \mathrm{~mL}$
$31.6 \mathrm{~mL}$
$36.1 \mathrm{~mL}$
$26.3 \mathrm{~mL}$
Solution
From the given data $\mathrm{P}_{\mathrm{H} 2}=1500 \times 0.80$
$=1200 \mathrm{~mm}$ of $\mathrm{Hg}=\frac{1200}{760}$ atmosphere $=1.58$
atmosphere If 's are volumes of gas dissolved by same volume of liquid, then from Henry's law
$\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}$; or $\frac{\mathrm{V}_{2}}{20}=\frac{1.58}{1.0}$
or $\mathrm{V}_{2}=31.60 \mathrm{~mL}$
*