H2S, a toxic gas with rotten egg like sm | Class 12 Chemistry Chapter Solutions, Solutions NCERT Solutions

Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter Solutions. This page offers a step-by-step solution to the specific question from Exercise 1, Question 6: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 6:

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
  • May 12, 2019

Actually,you had taken 0.0987,instead of 0.987


  • Science
  • Mar 21, 2019

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
  • Mar 08, 2019

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


  • ananthu
  • Mar 08, 2019

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


  • Gi
  • Feb 11, 2019

How is p 0.987?


  • Sudip
  • Jul 01, 2018

How to say that miles of H2S is 0.195 mol


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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....