H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at S.T.P. is 0.195 m, calculate Henry's law constant.

It is given that the solubility of H₂S in water at STP is 0.195m, i.e., 0.195 mol of H₂S is dissolved in1000 g of water.
     
           =$$\begin{gathered} \frac{1000\mathrm{g}}{18\mathrm{g} {\mathrm{mol}}^{-1}} \\ moles of water =55.56 mol \\ =\frac{moles of H_{2}S }{moles of H_{2}S+ moles of water} \\ =\frac{0.195}{0.195+55.56} = 0.0035 \end{gathered}$$

At STP pressure (P) = 0.987bar
According to henry's law p = k_H x

$$\begin{gathered} {\mathrm{K}}_{\mathrm{H}} = \frac{\mathrm{p}}{\mathrm{x}} \\ = \frac{0.987}{0.0035}\mathrm{bar} = 282 \mathrm{bar} \end{gathered}$$