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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    At\[25{}^\circ C,\]at 5% aqueous solution of glucose (molecular weight\[=180\,g\,mo{{l}^{-1}}\]) is isotonic with a 2% aqueous solution containing and unknown solute. What is the molecular weight of the unknown solute?

    A)  60                         

    B)         80

    C)  72                         

    D)         63

    E)  98             

    Correct Answer: C

    Solution :

    Since the two solutions are isotonic, they must have same concentrations in moles/litre. For glucose solution, concentration \[=5\text{ }g/100\text{ }c{{m}^{3}}\] (given) \[=50g/L\] \[\therefore \]  \[\frac{50}{180}=\frac{20}{M}\] Or           \[M=72\]           (\[\because \]Molar mass of glucose\[=180\text{ }g\text{ }mo{{l}^{-1}}\]) For unknown substance, concentration \[=2\text{ }g/100\text{ }c{{m}^{3}}\] (given) \[=20g/L=\frac{20}{M}mol/L\] \[\therefore \]  \[\frac{50}{180}=\frac{20}{M}\] or            \[M=72\]


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