Question

The boiling point of an aqueous urea solution is $101.04{}^{\circ }C$. The density of solution is $1.12g/ml$. Which of the following is /are correct statement(s) regarding this solution?
[Given: For water: $K_b=0.52Kkgmo{l}^{-1},K_f=1.86Kkgmo{l}^{-1}]$

• A. The molarity of the solution is $2M$
• B. The freezing point of the solution is $3.72{}^{\circ }C$
• C. If the solution is cooled to $-1.86{}^{\circ }C$, no ice will form
• D. At any temperature, the relative lowering of vapour pressure is $\left(\frac{9}{259}\right)$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The molarity of the solution is $2M$
C If the solution is cooled to $-1.86{}^{\circ }C$, no ice will form
D At any temperature, the relative lowering of vapour pressure is $\left(\frac{9}{259}\right)$

(a) $\text{Elevation in the boiling point}$
$\Delta T_b=101.04{}^{\circ }C-100{}^{\circ }C=1.04{}^{\circ }C$
We know,
$\Delta T_b=K_{b}m$
$1.04=0.52\times m\Rightarrow m=2$

So, 2 moles of urea is present in 1000 g solvent

Mass of urea $w_B=2\times 60=120g$
Mass of the solvent, $w_A=1000g$
Mass of the solution, $w_s=(1000+120)g=1120g/mL$
Density of the solution, $d_s=1.12g/mL$
Volume of the solution, $v_s=\frac{1120}{1.12}=1000mL$
$Molarity=\frac{2}{1000}\times 1000=2M$


(b) WE know,
$\Delta T_f=K_{f}m=1.86\times 2=3.72$
$\Rightarrow \Delta T_f=T_f^0-T_f=3.72{}^{\circ }C$
$\Rightarrow 0-T_f=3.72{}^{\circ }C$
$\Rightarrow T_f=-3.72{}^{\circ }C$

(c) Since freezing point of solution $\left(T_f\right)=-3.72{}^{\circ }C$. So above this temperature, at $-1.86{}^{\circ }C$, no ice will form.

(d) $\text{relative lowering of vapor pressure}$
$=\text{mole fraction of the solute}$
$=\frac{2}{2+\frac{1000}{18}}=\frac{2\times 18}{1036}=\frac{9}{259}$