Vapour pressure of water at 293 Kis 17.5 | 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 34: . 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 34:

Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in 450 g of water.

Answer:

Vapour pressure of water, p1° = 17.535 mm of Hg

Mass of glucose, w2 = 25 g

Mass of water, w1 = 450 g

We know that,

Molar mass of glucose (C6H12O6),

M2 = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1

Molar mass of water, M1 = 18 g mol - 1

Then, number of moles of glucose, n1 = 25/180 = 0.139 mol

And, number of moles of water, n2 =450/18 = 25 mol

Now, we know that,

(p1° - p°) / p1° =  n1 / n2 + n1

⇒ 17.535 - p°  / 17.535 =   0.139 / (0.139+25)

⇒ 17.535 - p1 = 0.097

⇒ p1 = 17.44 mm of Hg

Hence, the vapour pressure of water is 17.44 mm of Hg.


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Comments

  • anonymous
  • Apr 10, 2021

very good


  • Abhi
  • Jun 16, 2019

Did\'nt understand the last second step how does 0.097 comes? Please answer....


  • Aqeel Ahmad
  • May 30, 2019

Bole to jhakkas


  • shamim abbas laskar
  • Feb 13, 2019

thnx


  • Rashmika
  • Oct 25, 2018

Irrelevant answer .. Waste of time... Not at all satisfied...


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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 34: Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhe....