Calculate the mass of a non-volatile sol | 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 2, Question 18: . 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 18:

Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Answer:

Let the vapour pressure of pure octane be p10.

Then, the vapour pressure of the octane after dissolving the non-volatile solute is 80/100 p10 = 0.8 p10.

Molar mass of solute, M2 = 40 g mol - 1

Mass of octane, w1 = 114 g

Molar mass of octane, (C8H18), M1 = 8 × 12 + 18 × 1 = 114 g mol - 1

Applying the relation,

(p10 - p1) / p10    (w2 x M1 ) / (Mw1 )

(p10 - 0.8 p10) / p10    (w2 x 114 ) / (40  x 114 )

⇒ 0.2 p10 / p10   w2 / 40

⇒ 0.2 = w2 / 40

w2 = 8 g

Hence, the required mass of the solute is 8 g.


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Comments

  • Pankaj
  • May 24, 2019

Give the complete & correct solution, immediately


  • Tanya
  • May 04, 2019

This solution is quite easy but this is not correct. The correct answer for this question is 10.


  • Tanya
  • May 04, 2019

This is quite easy but correct answer is 10


  • Bhagya
  • Apr 25, 2019

For the answer to be 10 , it should be equated with X of solute i.e. ((w1/M1)/(w1/M1)+(w2/M2))


  • Gautam
  • Apr 22, 2019

Correct is 10


  • Neha
  • Mar 28, 2019

But answer is 10


  • Neha
  • Mar 28, 2019

Give the correct solution


  • Renuka
  • Mar 10, 2019

Answer is right


  • Malavika Manish
  • Mar 09, 2019

The answer is wrong ....the correct answer should be done like given below if vapour pressure of pure liquid is = Po 80 % of pure liquid Ps= 80×Po/100 = 0.8Po Ps =Po × Xsolute mass of solute = x gram mass of solvent = 114g Molar mass of solute= 40 g/mol Molar mass of solvent (octane C8H18) = 114g/mol Number of moles of solute = x/40 = 0.025x Number of moles of solvent = 114/114= 1 moles Mole fraction of solvent = 1/(1+0.025x) 0.8Po=Po×1/(1+0.025x) Cross multiply we get (1+0.025x))0.8Po= Po Divide by 0.8 Po we get 1+0.025x = 1.25 Subtract 1 both side we get 0.025x = 0.25 Now divide by 0.025 we get x = 10g


  • nikhil gupta
  • Jan 17, 2019

this answer in wrong instead the formula used will be p-ps/ps


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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 2 , Question 18: Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) which should be dissolved in 114....