Solutions Question 336
Question: A solution of urea (mol. mass $ 56gmo{l^{-1}} $ ) boils at $ 100.18{}^\circ C $ at the atmospheric pressure. If $ K_{f} $ and $ K_{b} $ for water are 1.86 and $ 0.512Kkgmo{l^{-1}} $ respectively, the above solution will freeze at
Options:
A) $ 0.654{}^\circ C $
B) $ -0.654{}^\circ C $
C) $ 6.54{}^\circ C $
D) $ -6.54{}^\circ C $
Show Answer
Answer:
Correct Answer: B
Solution:
As $ \Delta T_{f}=K_{f}m $
$ \Delta T_{b}=K_{b}.m $ Hence, we have $ m=\frac{\Delta T_{f}}{K_{f}}=\frac{\Delta T_{b}}{K_{b}} $ or $ \Delta T_{f}=\Delta T_{b}\frac{K_{f}}{K_{b}} $
$ [ \Delta T_{b}=100.18-100=0.18{}^\circ C ] $
$ =0.18\times \frac{1.86}{0.512}=0.654{}^\circ C $ As the freezing point of pure water is $ 0{}^\circ C, $
$ \Delta T_{f}=0-T_{f} $
$ 0.654=0-T_{f} $
$ \therefore T_{f}=-0.654 $ Thus the freezing point of solution will be $ -0.654{}^\circ C. $
