Thank you for reporting, we will resolve it shortly
Q.
$H_{2}S$ a toxic gas with rotten egg like smell, is used for qualitative analysis. If the solubility of $H_{2}S$ in water at $STP$ is $0.195\, m$, Henry’s law constant is
Solutions
Solution:
Solubility of $H_{2}S$ gas $= 0.195\, m$
$\therefore $ Moles of $H_{2}S = 0.195$, Mass of water $= 1000 \, g$
No. of moles of water $ = \frac{1000 \,g}{18 \,g\, mol^{-1}} = 55.55$ moles
$\therefore $ Mole fraction of $H_{2}S$ gas in the solution, (x)
$ = \frac{0.195}{0.195 + 55.55} = \frac{0.195}{55.745} = 0.0035$
Pressure at $STP = 0.987\,$ bar
Applying Henry’s law, $p (H_{2}S) = K_{H} \times x_{H_2S}$
or $K_{H} = \frac{P_{H_2S}}{x_{H_2S}} = \frac{0.987 \, bar}{0.0035} = 282\,$ bar