Question
A solution of urea $$\left( {mol.\,mass\,56\,g\,mo{l^{ - 1}}} \right)$$ boils at $${100.18^ \circ }C$$ at the atmospheric pressure. If $${K_f}$$ and $${K_b}$$ for water are $$1.86$$ and $$0.512\,K\,kg\,mo{l^{ - 1}}$$ respectively, the above solution will freeze at
A.
$${0.654^ \circ }C$$
B.
$$ - {0.654^ \circ }C$$
C.
$${6.54^ \circ }C$$
D.
$$ - {6.54^ \circ }C$$
Answer :
$$ - {0.654^ \circ }C$$
Solution :
$$\eqalign{
& {\text{As}}\,\,\Delta {T_f} = {K_f}m \cr
& \Delta {T_b} = {K_b}.\,m \cr
& {\text{Hence, we have}}\,m{\text{ = }}\frac{{\Delta {T_f}}}{{{K_f}}} = \frac{{\Delta {T_b}}}{{{K_b}}} \cr
& {\text{or}}\,\,\Delta {T_f} = \Delta {T_b}\frac{{{K_f}}}{{{K_b}}} \cr
& \Rightarrow \left[ {\Delta {T_b} = 100.18 - 100 = {{0.18}^ \circ }C} \right] \cr
& = 0.18 \times \frac{{1.86}}{{0.512}} \cr
& = {0.654^ \circ }C \cr} $$
As the freezing point of pure water is $${0^ \circ }C,$$
$$\eqalign{
& \Delta {T_f} = 0 - {T_f} \cr
& 0.654 = 0 - {T_f} \cr
& \therefore \,\,{T_f} = - 0.654 \cr} $$
Thus the freezing point of solution will be $$ - {\text{0}}{\text{.65}}{{\text{4}}^ \circ }C.\,$$