`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

Solution of `H_(2)S` gas =0.195m
`=0.195` mole in 1 kg of solvent
1 kg of solvent =1000g
`=(1000)/(18)=55.55` moles
`thereforex_(H_(2)S)=(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)`
`impliesK_(H)=(PH_(2)S)/(x_(H_(2)S))=(0.987)/(0.0035)=282` bar