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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    \[A\,\,0.5\,\,g/L\] solution of glucose is found to be isotonic with a \[2.5\,\,g/L\] solution of an organic compound. What will be the molecular weight of that organic compound?

    A) \[300\]                                 

    B)        \[600\]

    C)        \[900\]                                 

    D)        \[200\]

    Correct Answer: C

    Solution :

    Molarity of glucose solution                 \[=\frac{0.5\,\,g}{180\times 1L}=0.00277\,\,M\] Isotonic solution have equal molar concentrations, hence molarity of organic compound solution\[=0.00277\,\,M\] \[\therefore \]Molecular weight of compound                                       \[=\frac{2.5}{0.00277\times 1}=900\]


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