Calculate (a) molality (b) molarit | 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 5: . 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 5:

Calculate

(a) molality

(b) molarity and

(c) mole

fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.

Answer:

(a) Molar mass of KI = 39 + 127 = 166 g mol - 1

20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.

That is,

20 g of KI is present in (100 - 20) g of water = 80 g of water

Therefore, molality of the solution = Moles of KI / Mass of water in kg

= 20/166 / 0.08 m

= 1.506 m

= 1.51 m (approximately)

 

(b) It is given that the density of the solution = 1.202 g mL - 1

∴Volume of 100 g solution = Mass /  Density

= 100g / 1.202g mL-1

= 83.19 mL

= 83.19 × 10 - 3 L

Therefore, molarity of the solution = 20/166 mol 83.19 × 10 - 3 L

= 1.45 M

 

(c) Moles of KI = 20/166 = 0.12 mol

Moles of water = 80/18 = 4.44 mol

Therefore, mole fraction of KI = Moles of KI / (Moles of KI + Moles of water)

= 0.12 / (0.12+4.44)

= 0.0263


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Comments

  • Gaganpreet
  • Jun 17, 2019

Thanks it was easy to understand with ur explanation


  • Mriganka Basu
  • Jun 02, 2019

Thanks, your solutions were quite useful!!


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  • May 27, 2019

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  • May 27, 2019

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  • Naincy
  • May 22, 2019

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  • May 02, 2019

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  • Shubham pandey
  • Apr 20, 2019

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  • Apr 04, 2019

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  • Mar 28, 2019

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  • Nixon MJ
  • Mar 09, 2019

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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 5: Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (m....