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 | 4th 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
     ? 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

Answered by  | 13th Jun, 2011, 03:27: PM

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