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NCERT Class XII Chemistry
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Question : 47 of 53
Marks: +1, -0
H2S\mathrm{H_2S} a toxic gas with rotten egg like smell, is used for qualitative analysis. If the solubility of H2S\mathrm{H_2S} in water at STP is 0.195 m, calculate Henry’s law constant.
Solution:  
Solubility of H2S gas=0.195 m\mathrm{H_2S}\ \text{gas}=0.195\ \text{m}
∴  \therefore\; Moles of H2S=0.195,\mathrm{H_2S}=0.195, Mass of water =1000 g=1000\ \mathrm{g}
No. of moles of water 1000 g18 g mol−1\frac{1000\ \mathrm{g}}{18\ \mathrm{g}\ \mathrm{mol}^{-1}}
=55.55 moles=55.55\ \text{moles}
∴ Mole fraction of H2S\mathrm{H_2S} gas in the solution (x)(x)
=0.1950.195+55.55=0.19555.745=0.0035=\frac{0.195}{0.195+55.55}= \frac{0.195}{55.745}=0.0035
Pressure at STP = 1 bar
Applying Henry's law, P( H2S )=KH×xH2SP_{(\ \mathrm{H_2S}\ )}=K_{H} \times x_{\mathrm{H_2S}}
or, KH=PH2SxH2SK_{H}=\frac{P_{\mathrm{H_2S}}}{x_{\mathrm{H_2S}}}
=1 bar0.0035=285.7 bar= \frac{1\ \text{bar}}{0.0035}=285.7\ \text{bar}
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