2. Sample Papers
  3. KVPY

KVPY Sample Paper KVPY Stream-SX Model Paper-11

  • 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. \[({{K}_{f}}=1.86K\,\,\text{Kg}\,\,\text{mo}{{\text{l}}^{-1}},\,\,{{K}_{b}}=0.52K\,\,\text{Kg}\,\,\text{mo}{{\text{l}}^{-1}})\]

    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) \[=100+\Delta \,{{T}_{b}}-(-\Delta \,{{T}_{f}})\]
    \[104=100+\Delta \,{{T}_{b}}+\Delta \,{{T}_{f}}=100+\text{molality}\times {{\text{K}}_{\text{b}}}+\text{molality}\times {{\text{K}}_{f}}\]\[=100+\text{molality}\,\text{(0}\text{.52+186)}\]
    \[\therefore \]      Molality \[=\frac{104-100}{2.38}=\frac{4}{2.38}=1.68\,m\]
    And molality \[=\frac{\text{moles}\times \text{1000}}{{{W}_{gm}}_{\,(solvent)}};\] \[\text{1}\text{.68}=\frac{\text{moles}\times \text{1000}}{500}\]
    \[\therefore \]      Moles of solute \[=\frac{1.68\times 500}{1000}=0.84\]moles


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