2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The difference between the boiling point and freezing point of an aqueous solution containing sucrose (molecular wt = 342 g\[mo{{l}^{-1}}\]) in 100 g of water is\[105.0{}^\circ C\]. If\[{{k}_{f}}\]and\[{{k}_{b}}\]of water are 1.86 and\[0.51\text{ }K\text{ }kg\text{ }mo{{l}^{-1}}\] espectively, the weight of sucrose in the solution is about

    A)  34.2 g                  

    B) 342 g

    C) 7.2 g                      

    D)        72 g

    E) 68.4 g

    Correct Answer: D

    Solution :

    Given, \[{{T}_{b}}-{{T}_{f}}=105.0{}^\circ C\] \[\Rightarrow \]\[(100+\Delta {{T}_{b}})-(0-\Delta {{T}_{f}})=105{}^\circ C\] \[\therefore \]                  \[\Delta {{T}_{b}}+\Delta {{T}_{f}}=5\] \[\Delta {{T}_{b}}+\Delta {{T}_{f}}=({{k}_{b}}+{{k}_{f}})\times m\]\[(m=molality)\] \[\Rightarrow \]               \[5=(1.86+0.51)\times \frac{w\times 1000}{342\times 100}\] \[\therefore \]  \[w=\frac{1710}{23.7}=72\,g\]


You need to login to perform this action.
You will be redirected in 3 sec spinner