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A solution of urea (mol. mass 56 g mol$^{-1}$) 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 mol$^{-1}$ respectively, the above solution will freeze at :-
A
$-6.54^\circ C$
B
$-0.654^\circ C$
C
$6.54^\circ C$
D
$0.654^\circ C$
Detailed Solution
Elevation of boiling point: $\Delta T_b = 100.18 - 100 = 0.18$ K
$\Delta T_b = K_b m \Rightarrow m = \dfrac{0.18}{0.512} = 0.3516$ mol kg$^{-1}$
Depression of freezing point: $\Delta T_f = K_f m = 1.86 \times 0.3516$
$\Delta T_f = 0.654$ K
Shortcut: $\dfrac{\Delta T_f}{\Delta T_b} = \dfrac{K_f}{K_b} \Rightarrow \Delta T_f = 0.18 \times \dfrac{1.86}{0.512} = 0.654$ K
Freezing point of the solution $= 0 - 0.654 = -0.654^\circ C$
The molar mass of urea is not needed for this calculation.
$\Delta T_b = K_b m \Rightarrow m = \dfrac{0.18}{0.512} = 0.3516$ mol kg$^{-1}$
Depression of freezing point: $\Delta T_f = K_f m = 1.86 \times 0.3516$
$\Delta T_f = 0.654$ K
Shortcut: $\dfrac{\Delta T_f}{\Delta T_b} = \dfrac{K_f}{K_b} \Rightarrow \Delta T_f = 0.18 \times \dfrac{1.86}{0.512} = 0.654$ K
Freezing point of the solution $= 0 - 0.654 = -0.654^\circ C$
The molar mass of urea is not needed for this calculation.