The solubility of `H_2S`gas in water at STP.195m .Calculate its Henry's law constant at STP

Explanation

Solubility of `H_2S` at STP = 0.195m,
i.e., 0.195 moles of `H_2S` is dissolved in 1000g (1kg) of water
Number of moles in 1000g `H_2O` = 1000/18 = 55.55
`therefore` Mole fraction of `H_2S` in the solution (x) =` 0.195/(0.195+55.55)` = `0.195/55.745`= 0.0035
STP pressure =1 bar
Applying Henry.s law,`P_(H_2S)= K_H xx x_(H_2S) therefore K_H of H_2S = P_(H_2S)/x_(H2S)` = 1 bar/0.0035 = 285.7 bar