When a gas is bubbled through water at 298K, a very dilute solution of the gas is obtained. Henry's law constant for the gas at 298K is 100kbar. If the gas exerts a partial pressure of 1 bar, the number of millimoles of the gas dissolved in one litre of water is(round off the answer to two digits after the decimal point)

$\begin{array}{r}{\mathrm{k}}_{\mathrm{H}}=100\mathrm{kbar},\mathrm{p}=1\mathrm{bar}\\ \mathrm{p}={\mathrm{k}}_{\mathrm{H}}\times \text{ molefraction }\end{array}$

Mole fraction $=\frac{\mathrm{p}}{{\mathrm{k}}_{\mathrm{H}}}=\frac{1}{100\times {10}^3}={10}^{-5}$

Mole fraction $=\frac{\text{ Mole of gas }}{\text{ Total moles }}$

$\text{ Moles of water }=\frac{\text{ Weight of water }}{\text{ Molecular weight of water }}$

$\text{ Weight of water }=1000\mathrm{g}(\because 1000\mathrm{mL}=1000\mathrm{g})$

$\begin{array}{l}\text{ Moles }=\frac{1000}{18}=55.5\\ \text{ Mole fraction }={10}^{-5}=\frac{x}{55.5+x}\end{array}$

As 55.5 >>> x, thus neglecting x from denominator

${10}^{-5}=\frac{x}{55.5}\Rightarrow x=55.5\times {10}^{-5}\text{ moles }$ or $$\begin{gathered} 55.5\times {10}^{-2}\text{ millimoles }=0.555 \mathrm{millimoles} \\ \approx 0.56 \mathrm{millimoles}. \end{gathered}$$