`H_(2)S` a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of `H_(2)S` in water at STP is `0.195` m. Calculate Henry's law constant.

Explanation

Solubility of `H_(2)S` gas `=0.195 m`
`=0.195` mole in 1 kg of the solvent (water)
1 kg of the solvent (water) `= 1000 gh = (1000g)/(18gmol^(-1)) = 55.55` moles
Mole fraction of `H_(2)S` gas in the solution `(x) = (0.195)/( 0.195 + 55.55) = (0.195)/(55.745) = 0.0035`
Pressure at STP `=0.987` bar
Applying Henry.s law `P_(H_(2)S) = K_(H) xx x _(H_(2)S)`
` K _(H) = (P _(H_(2)S))/(x_(H_(2)S))= (0. 987 b ar)/(0.0035) = 282` bar