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KVPY Sample Paper KVPY Stream-SX Model Paper-31

  • question_answer
    How many moles of sucrose should be dissolved in 500 gms of water so as to get a solution which has a difference of \[104{}^\circ C\] between boiling point and freezing point. \[({{K}_{f}}=1.86\,\,K\,\,kg\,\,mo{{l}^{-1}},\] \[{{K}_{b}}=0.52\,\,K\,\,Kg\,\,mo{{l}^{-1}})\]

    A) 1.68     

    B) 3.36

    C) 8.40     

    D) 0.840

    Correct Answer: D

    Solution :

    Boiling point of solution =boiling point \[+\Delta {{T}_{b}}=100+\Delta {{T}_{b}}\]
    Freezing point of solution = freezing
    Point \[-\,\Delta {{T}_{f}}\]                    \[=0-\Delta {{T}_{f}}\]
    Difference in temperature (given) \[=100+\Delta {{T}_{b}}-\,(-\,\Delta {{T}_{f}})\]
    \[104=100+\Delta {{T}_{b}}+\Delta {{T}_{f}}\]
    =100 + molality \[\times \,\,{{K}_{b}}\]+molality \[\times \,\,{{K}_{f}}\]
    =100+molality (0.52+1.86)
    \[\therefore \] Molality \[=\frac{104-100}{2.38}=\frac{4}{2.38}=1.68\,\,m\]and molality \[=\frac{Moles\times 1000}{{{W}_{gm\,\,(solvant)}}}\,\,;\]
    \[1.68=\frac{Moles\times 1000}{500}\]
    \[\therefore \]Moles of solute \[=\frac{1.68\times 500}{1000}\]= 0.84 moles


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