H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry's law constant.
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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9 Comment(s) on this Question
The Henry's law analysis can be done in many possible units including atm. It is done in the units of KBar here, as per NCERT. Yes 1atm=.0987bar
Sir there is an error 0.0987 instead of 0.987
Sir there is an error 0.0987 instead of 0.987
Actually,you had taken 0.0987,instead of 0.987
Pressure at STP is 1 atm And 1 atm = 1.01325 bar Acc. to Henryâs Law p= KHx â KH = p / x KH = 1.0325/0.0035 = 289.5 bar This is the correct answer
I think there is an error in the second last step, insted of .987, .0987 is used
molality is .195 means , .195 moles of solute in 1000 g water
How is p 0.987?
How to say that miles of H2S is 0.195 mol
- NCERT Chapter