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Solutions

Question

H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at S.T.P. is 0.195 m, calculate Henry's law constant.

Solution

It is given that the solubility of H₂S in water at STP is 0.195m, i.e., 0.195 mol of H₂S is dissolved in1000 g of water.

= $\frac{1000 g}{18 g {mol}^{- 1}} m o l e s o f w a t e r = 55.56 m o l = \frac{m o l e s o f H_{2} S}{m o l e s o f H_{2} S + m o l e s o f w a t e r} = \frac{0.195}{0.195 + 55.56} = 0.0035$

At STP pressure (P) = 0.987bar
According to henry's law p = k_Hx

$K_{H} = \frac{p}{x} = \frac{0.987}{0.0035} bar = 282 bar$

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Solutions

H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at S.T.P. is 0.195 m, calculate Henry's law constant.

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For Daily Free Study Material Join wiredfaculty

Solutions

H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at S.T.P. is 0.195 m, calculate Henry's law constant.

Some More Questions From Solutions Chapter

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at S.T.P. is 0.195 m, calculate Henry's law constant.