Question

Calculate henry's constant if the solubility of H2S in water at stp is 0.195m?

Answer 1

i.e'0.195 mol of H2S is dissolved in 1000g
moles of water =1000 g/ 18 g mol-1
              =55.56 mol
mole fraction of H2S,x =moles of H2S/moles of H2S + moles of water
                                           =0.195/0.195 +55.56 =0.0035
At STP ,pressure {p] =0.987 bar 
According to Henry's law;
   p=Khx
   Kh =p/x
    0.987/0.0035 bar 
                              = 282 bar