At $20^{\circ} \mathrm{C}$ and $1.00$ atm partial pressure of $\mathrm{H}_{2}, 18$ $\mathrm{mL}$ of…

At $20^{\circ} \mathrm{C}$ and $1.00$ atm partial pressure of $\mathrm{H}_{2}, 18$ $\mathrm{mL}$ of $\mathrm{H}_{2}$ (STP) dissolves in 1 L of water. If $2 \mathrm{~L}$ of water is exposed to a gaseous mixture having a total pressure of 1425 torr (excluding the vapour pressure of water) and containing $80 \% \mathrm{H}_{2}$ by volume, the volume of $\mathrm{H}_{2}$ (STP) dissolved is
  1. $27 \mathrm{~mL}$
  2. $54 \mathrm{~mL}$
  3. $33.75 \mathrm{~mL}$
  4. $67.50 \mathrm{~mL}$

Solution

$\mathrm{V}_{\text {dissolved }}=\mathrm{KP}_{\mathrm{H} 2}$
$\mathrm{P}_{\mathrm{H} 2}$ in the mixture $=\mathrm{X}_{\mathrm{H} 2} \mathrm{P}=\frac{80 \times 1425}{100}=1140$
torr $=\frac{1140}{760}=1.5 \mathrm{~atm}$
Hence, $V_{\text {dissolved }}$ in $2 \mathrm{~L}$ of water $=2 \times \mathrm{K} \times \mathrm{P}_{\mathrm{H} 2}$
$=2 \times 18 \times 1.5=54 \mathrm{~mL}$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more SOLUTIONS questions on Aicharya