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.
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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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....
Comments
Difference in temp in what ever measure will be the same.no need of adding 273 and subtracting it again
Very easy method Thank you
Can u show me another method
Good
It's not clear plzz take different method
Write anwers
No it's not
But the answer on ncert text book is 1.86g