H2S, a toxic gas with rotten egg like sm | Class 12 Chemistry Chapter Solutions, Solutions NCERT 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 STP is 0.195 m, calculate Henry's law constant.

Answer:

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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Comments

  • Komal
  • 2019-05-12 17:55:07

Actually,you had taken 0.0987,instead of 0.987


  • Science
  • 2019-03-21 11:54:52

Pressure at STP is 1 atm And 1 atm = 1.01325 bar Acc. to Henry’s Law p= KHx ⇒ KH = p / x KH = 1.0325/0.0035 = 289.5 bar This is the correct answer


  • ananthu
  • 2019-03-08 22:54:37

I think there is an error in the second last step, insted of .987, .0987 is used


  • ananthu
  • 2019-03-08 22:53:01

molality is .195 means , .195 moles of solute in 1000 g water


  • Gi
  • 2019-02-11 11:04:21

How is p 0.987?


  • Sudip
  • 2018-07-01 15:23:18

How to say that miles of H2S is 0.195 mol


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 6: H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility....