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.
Solution
Normal boiling point of water = 100
C
Elevation of boiling point, ∆Tb = 100 − 99.63 = 0.37
Mass of water w1= 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
Use the formula
w2= 121.67 g
Answer mass of added sucrose = 121.67 g
Should the temperature by converted to kelvin? as it is the SI unit...
ReplyDeleteNo. We only take change in boiling pt.(temperature). So even if we vonveet it and find the change. The valuv remains the same
DeleteThough we convert it to Kelvin we still get the same difference as in Celsius scale. Why did u take 3.37 k instead of 0.37
ReplyDeleteIt's a mistake it shoulb be 0.37
DeleteRegards,
ReplyDelete@Admin
Please explain more about Vapour pressure of water
The molar mass of sucrose is 330 not 342...
ReplyDeleteplz. correct it
342 is right..
DeleteThe molar mass of sucrose is 330 not 342...
ReplyDeleteplz. correct it
It's not clear plzz take different method else
ReplyDeleteOn ncert answer of this question is given 1.86g
ReplyDeleteWhish one is correct answer (121.67g or 1.86g) ????
ReplyDelete