1. Tardigrade
  2. Question
  3. Chemistry
  4. An aqueous solution boils at 100.2° C . At which temperature this will freeze. (Kb=0.5° C/m,Kf=1.9° C/m)

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Q. An aqueous solution boils at $ 100.2{}^\circ C $ . At which temperature this will freeze. $ ({{K}_{b}}=0.5{}^\circ C/m,{{K}_{f}}=1.9{}^\circ C/m) $

Rajasthan PETRajasthan PET 2009

Solution:

$ \because $ $ {{T}_{b}}=100.2{}^\circ C $
$ \therefore $ $ \Delta {{T}_{b}}=100.2-100=0.2{}^\circ C $
$ \Delta {{T}_{b}}={{K}_{b}}\times m $
$ m=\frac{\Delta {{T}_{b}}}{{{K}_{b}}}=\frac{0.2}{0.5}=0.4 $
$ \Delta {{T}_{f}}={{K}_{f}}\times m=1.9\times 0.4 $
$ =0.76{}^\circ C $
$ \Delta {{T}_{f}}={{T}_{o}}-\Delta {{T}_{f}}=0-0.76 $
$ =-0.76{}^\circ C $