how much grams urea should be dissolved in 500g of water to get the boiling point of solution at 101.5°c
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1
Answer:
Explanation:
Boiling point of solution = 100.18^Given : Boiling point of solution =100.18
o
C
K
b
=0.512Kkgmol
−1
K
f
=1.86Kkgmol
−1
ΔT Boiling =100.18
o
−100
o
=0.18
ΔT=K
b
×molality
∴ molality =
0.512
0.18
Now, ΔT
freezing
=K
f
×molality
=1.86×
0.512
0.18
=0.6oC Kb = 0.512 K kg mol^-1
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