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It is given that the solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water.
Moles of water = 1000g / 18g 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
= 0.0987 / 0.0035 bar
= 282 bar
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