If the solubility of a gas, A in water at N.T.P. and $0.95 \mathrm{bar}$ is $0.15 \mathrm{m}$, The Henry law constant will be

Molality of A is 0.15 m indicates that 0.15 moles present in 1 kg of solvent (1000 g of solvent).

Here, solvent is water and moles of water in 1000 g is calculated as:

$\mathrm{Moles}=\frac{1000 \mathrm{g}}{18 \mathrm{g}/\mathrm{mol}}=55.55 \mathrm{moles}$

Mole fraction of solute is calculated as:

$$\begin{gathered} \mathrm{Mole} \mathrm{fraction} \mathrm{of} \mathrm{solute} \left({\mathrm{X}}_{\mathrm{A}}\right)=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{moles} \mathrm{of} \mathrm{solute}+\mathrm{moles} \mathrm{of} \mathrm{solvent}} \\ \mathrm{Mole} \mathrm{fraction} \mathrm{of} \mathrm{solute} \left({\mathrm{X}}_{\mathrm{A}}\right)=\frac{0.15 \mathrm{mole}}{0.15 \mathrm{mole}+55.55 \mathrm{mole}}=0.0027 \end{gathered}$$

Henry's constant is calculated as:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{A}}={\mathrm{K}}_{\mathrm{H}}\times {\mathrm{X}}_{\mathrm{A}} \\ {\mathrm{K}}_{\mathrm{H}}=\frac{0.95}{0.0027}=351.85 \mathrm{bar} \end{gathered}$$