Vapour pressure of water at 293 Kis 17.5 | Class 12 Chemistry Chapter Solutions, Solutions NCERT Solutions

Question:

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
  • 2021-04-10 14:50:18

very good


  • Abhi
  • 2019-06-16 11:29:11

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


  • Aqeel Ahmad
  • 2019-05-30 00:52:25

Bole to jhakkas


  • shamim abbas laskar
  • 2019-02-13 19:23:39

thnx


  • Rashmika
  • 2018-10-25 19:49:23

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


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