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NCERT Class XII Chemistry
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Question : 47
Total: 53
H2S a toxic gas with rotten egg like smell, is used for qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.
Solution:  
Solubility of H2S gas=0.195 m
∴‌ Moles of H2S=0.195, Mass of water =1000 g
No. of moles of water =
1000g
18g mol−1

=55.55 moles
∴ Mole fraction of H2S gas in the solution (x)
=
0.195
0.195+55.55
=
0.195
55.745
=0.0035
Pressure at STP = 1 bar
Applying Henry's law, P ( H2S )=KH×x H2S
or, KH= P H2 Sx H2 S
=
1bar
0.0035
=285.7 bar
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