(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269·15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273·15 K.
Given :
(Molar mass of sucrose = 342 g mol-1)
(Molar mass of glucose = 180 g mol-1)
(b) Define the following terms :
(i) Molality (m)
(ii) Abnormal molar mass

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

(a) Molor mass of sucrose = 342g mol-1
   Molar mass of glucose = 180 g mol-1
 Tf  = ?
   \Delta T_{f}= K_{f}m
   \Delta T_{f}= T_{f^{0}}-T_{f}
  T_{f^{0}}-T_{f}= K_{f}m
273\cdot 15-269\cdot 15= K_{f}\times \frac{10}{342}\times \frac{1000}{100}
K_{f}= \frac{4\times 342}{10\times 10}
         = 13\cdot 68Kmol^{-1}
273\cdot 15-T_{f} = 13\cdot 68\times \frac{10}{180}\times \frac{1000}{100}
273\cdot 15-7\cdot 6= T_{f}
T_{f}= 273\cdot 15-7\cdot 6=265\cdot 55 K
(b) (i) Molality (m)
as the number of moles of solute present per kg of solvent unit of m is mol kg-1 or molal
(ii) Abnormal molar mass - If the molar mass of the solute calculated by using the colligative properly is formed to be different than its normal molar mass than the solute is said to exhibit/possess abnormal molar mass.
 

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