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Grade 12Organic Chemistry

H2S ,a toxic gas with rotten egg like smell,is used for qualitative analysis. If the solubility of H2S in water at STP is 0.195m, calculate Henry's law constant

Profile image of Nayana
7 Years agoGrade 12
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1 Answer

Profile image of Arun
7 Years ago

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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