Question

A solution of urea (mol. Mass 56 gmol-1) boils at 100.18C at the atmospheric pressure. If $K_{f}$ and $K_{b}$ for water are 1.86 and 0.52 $K kg mol^{-1}$, respectively, the above solution will freeze at:
A. -6.54 ${ } ^{0} C$
B. 6.04 ${ } ^{0} C$
C. 0.654 ${ } ^{0} C$
D. -0.654 ${ } ^{0} C$

Answer 1

Molar mass of urea is given. We can use the formulae of molality i.e
Molality = $\dfrac{\Delta T}{K_{b}}$. The solvent here is water, and the values of $K_{f}$ and $K_{b}$ are given to us, which will help us in determining the boiling and freezing point of the given solution that will solve the question required.

Complete step by step answer:
In order to find the freezing temperature of the urea, we need to first find the freezing point and elevation point of the solution. In order to do so
We start by calculating the $\Delta T_{f}$ and $\Delta T_{b}$ as follows:
Depression is a freezing point ($\Delta T_{f}$) is:
$\Delta T_{f}$ = $K_{f}$ × molality of the solution
The elevation in boiling point() is
$\Delta T_{b}$ = $K_{b}$ × molality of the solution
Hence the formula we are going to use is
$\Delta T_{f} \times \Delta T_{b}=K_{b} \times K_{f}$
​Given that
$\Rightarrow \Delta T_{b}=T_{2}-T_{1}$
$\Rightarrow =100.18-100$
$=0.18^{0} \mathrm{C}$
The freezing point($K_{f}$)for water = 1.86 K kg mol−1
The elevation point ($K_{b}$ ) for water = 0.52 K kg mol−1
∴ 0.18 ${\Delta T_{f}}=0.5121 .86$
or $\Delta T_{f}$ = 0.5121.86×0.18 = 0.6539≈0.65
${\Delta T_{f}=T_{1}-T_{2}} \\\ 0.654= 0{ }^{0} C-T_{2}$
$\therefore T_{2}=-0.654^{0} C$
$\left(T_{2}\right) \rightarrow$ The freezing point of aqueous urea solution is.
Hence the freezing point for solution is $T_{2}=-0.654^{0} C$.

**Therefore option D is correct.

Note:**
Pay heed to the solvent that is given in the question. As for this question, it's water. Know which formula to use by learning the theory correctly. In this question, the formula used is the elevation in boiling point and depression in freezing point.