Solubility of a gas in water at STP is 0.2m. Henry's constant of the gas is equal to 

${\mathrm{p}}_{\mathrm{A}}={\mathrm{k}}_{\mathrm{A}}{\mathrm{X}}_{\mathrm{A}}\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(1\right)$

Given,

Molality of the solution = 0.2m

Number of moles of solute = 0.2

Weight of water = 1000g

Number of moles of water =$\frac{1000}{18}$

At STP, ${\mathrm{p}}_{\mathrm{A}}=1 \mathrm{bar}$

$$\begin{gathered} {\mathrm{X}}_{\mathrm{A}}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}; \\ {\mathrm{n}}_2<<<{\mathrm{n}}_1 \\ \therefore {\mathrm{X}}_{\mathrm{A}}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}; \\ \mathrm{Substituting} \mathrm{these} \mathrm{values} \mathrm{in} \left(1\right) \\ 1={\mathrm{k}}_{\mathrm{H}}\times \frac{0.2}{\left({\displaystyle \frac{1000}{18}}\right)} \\ {\mathrm{k}}_{\mathrm{H}}=\frac{1000}{18\times 0.2}=278 \mathrm{bar} \end{gathered}$$