1. Tardigrade
  2. Question
  3. Chemistry
  4. How many gram of sucrose (mol. wt. = 342) should be dissolved in 100 g water in order to produce a solution with a 105.0° C difference between the freezing point and boiling temperature? [kf=1.860° C text/m, kb=0.151° C text/m]

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Q. How many gram of sucrose (mol. wt. = 342) should be dissolved in 100 g water in order to produce a solution with a $ 105.0{}^\circ C $ difference between the freezing point and boiling temperature? $ [{{k}_{f}}=1.860{}^\circ C\text{/}m,\,{{k}_{b}}=0.151{}^\circ C\text{/}m] $

CMC MedicalCMC Medical 2014

Solution:

Boiling point $ ({{T}_{b}})=100+\Delta {{T}_{b}}=100+{{k}_{b}}m $ Freezing point $ ({{T}_{f}})=0-\Delta {{T}_{f}}=-{{k}_{f}}m $ $ {{T}_{b}}-{{T}_{f}}=(100+{{k}_{b}}m)-(-{{k}_{f}}m) $ $ 105=100+0.51\,m+1.86\,m $ $ 2.37\,m=5 $ or $ m=\frac{5}{2.37} $ $ =2.11 $ $ \therefore $ Weight of sucrose to be dissolved in 100g water $ =\frac{2.11\times 342}{1000}\times 100 $ $ =72\,g $