RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

  • question_answer
    Elevation in boiling point of an aqueous urea solution is\[0.52{}^\circ C[{{K}_{b}}=0.52{}^\circ mo{{l}^{-1}}kg].\]Hence, mole fraction of urea in this solution is

    A)  0.982           

    B)  0.567

    C)  0.943           

    D)  0.018

    Correct Answer: D

    Solution :

     \[\because \] \[\Delta {{T}_{b}}=molality\times {{K}_{b}}\] \[\because \] \[0.52=m\times 0.52\] or   molality \[=1\text{ }mol\text{ }k{{g}^{-1}}\] \[\therefore \] No. of moles of urea = 1 mol No. of moles of water\[=\frac{1000}{18}=55.55\text{ }mol\] Mole fraction of urea \[=\frac{moles\text{ }of\text{ }urea}{moles\text{ }of\text{ }urea\text{ }+\text{ }moles\text{ }of\text{ }water}=0.018\] \[=\frac{1}{1+55.55}\]


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