NEET Class neet Answered
A 5 percent solution by mass of cane sugar in water has freezing point of 271k and freezing point of pure water is 273.15k the freezing point of a 5percent by mass of glucose in water is
Asked by pritumanojsingh9617935026 | 31 Mar, 2020, 12:11: 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 Ramandeep | 31 Mar, 2020, 01:33: PM
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