Boiling point of water at 750 mm Hg is 9 | Class 12 Chemistry Chapter Solutions, Solutions NCERT Solutions

Question:

Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.Molal elevation constant for water is 0.52 K kg mol-1.

Answer:

Here, elevation of boiling point ΔTb= (100 + 273) - (99.63 + 273)

= 0.37 K

Mass of water, wl = 500 g

Molar mass of sucrose (C12H22O11),

M2= 11 × 12 + 22 × 1 + 11 × 16 = 342 g mol - 1

Molal elevation constant, Kb= 0.52 K kg mol - 1

We know that:

= (0.37 x 342 x 500) /  (0.52 x 1000)

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.


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Comments

  • Tr
  • 2019-10-14 10:45:26

Difference in temp in what ever measure will be the same.no need of adding 273 and subtracting it again


  • Harshitha
  • 2019-08-29 21:26:17

Very easy method Thank you


  • Hasnehena khatun
  • 2019-05-30 19:06:50

Can u show me another method


  • munazza
  • 2019-05-29 21:12:35

Good


  • Shameema
  • 2019-04-25 18:41:01

It's not clear plzz take different method


  • Himanshu
  • 2018-05-30 15:34:28

Write anwers


  • Shambala
  • 2018-05-05 12:43:33

No it's not


  • Tawseef Manzoor
  • 2018-04-30 09:40:15

But the answer on ncert text book is 1.86g


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 1 , Question 10: Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of wate....