At 298 K , Henry's law constant of a gas ‘A’ dissolved in water is $1.013\times {10}^{-2} \mathrm{Kbar}$. Number of moles of ‘A’ dissolved in one kg of water at 1 atm and 298 K will be 

According to Henry's law

${\mathrm{p}}_{\mathrm{A}}={\mathrm{K}}_{\mathrm{H}}{\mathrm{X}}_{\mathrm{A}}$

${\mathrm{p}}_{\mathrm{A}}={\mathrm{K}}_{\mathrm{H}}\times \frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}}$

${\mathrm{n}}_{\mathrm{A}}<<<{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}$

$\therefore {\mathrm{n}}_{\mathrm{A}} \mathrm{in} \mathrm{the} \mathrm{denominator} \mathrm{can} \mathrm{be} \mathrm{neglected}$

${\mathrm{p}}_{\mathrm{A}}={\mathrm{K}}_{\mathrm{H}}\times \frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}}$

$$\begin{gathered} \mathrm{Given} {\mathrm{p}}_{\mathrm{A}}=1 \mathrm{atm}=1.013 \mathrm{bar} \\ {\mathrm{K}}_{\mathrm{H}}=1.013\times {10}^{-2} \mathrm{Kbar}=1.013\times 10 \mathrm{bar} \end{gathered}$$

${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=\frac{1000}{18}$

$$\begin{gathered} {\mathrm{n}}_{\mathrm{A}}=\frac{{\mathrm{p}}_{\mathrm{A}}\times {\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}}{{\mathrm{K}}_{\mathrm{H}}} \\ {\mathrm{n}}_{\mathrm{A}}=\frac{1.013\times 1000}{1.013\times 10\times 18}=5.55 \end{gathered}$$