CBSE Class 12-science Answered
A 5% solution ( by mass ) of cane sugar in water has freezing point of 271 K . calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K ?
Asked by sumit satpaty | 04 Jun, 2011, 05:22: PM
Expert Answer
For cane sugar ? Tf =273.15 -271.0 =2.15 K
Thus Kf = (? Tf x MB X WA)/( WB X1000)
=2.15 X 342 X100/(5 X1000)
=14.71 K Kg mol-1
For glucose solution
? Tf =Kf xWB X1000/(MB XWA)
= 14.71 X1000 X 5 /(100 X180)
=4.085
Therefore freezing point ofv 5% glucose solution is =273.15 -4.085=269.07
Answered by | 13 Jun, 2011, 03:27: PM
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