Calculate the molal elevation constant, kb for water and the boiling point of 0.1 molal urea solution. Latent heat of vaporisation of water is 9.72 kcal mol−1 at 373.15 K.
A
Kb=0.515Kkgmol−1,Tb=373.20K
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Kb=0.515Kkgmol−1,Tb=370.20K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Kb=0.565Kkgmol−1,Tb=373.20K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Kb=5.15Kkgmol−1,Tb=370.20K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is AKb=0.515Kkgmol−1,Tb=373.20K The expression for the molal elevation constant is kb=0.002(Tob)2Lv
Here, Tob is the boiling point of pure solvent and Lv is the latent heat of vaporization in cal/g of the pure solvent. It is equal to 9.72kcalmol−1or540cal/g.
∴kb=0.002×(373.15)2540=0.515Kkgmol−1
Th depression in freezing point is ΔTb=kb×m=0.515×0.1=0.0515