Question
Sat June 04, 2011 By: Sumit Satpaty
 

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 ?

Expert Reply
Mon June 13, 2011
 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
     ? T =Kf xW 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
Related Questions
Tue October 17, 2017

what is edema?

Home Work Help