Question

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

• A. The molarity of solution is 2 M
• B. The freezing point of 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)

At any temperature, the relative lowering of vapour pressure is $\left(\frac{9}{259}\right)$

A) $\Delta T_b=K_{b}m$
Here,
$\Delta T_b$ is the elevation in boiling point $=101.04{}^{\circ }C-100{}^{\circ }C=1.04{}^{\circ }C$
$k_b$ is molal elevation in boling point constant and $m$ is the molality
Hence,
$1.04=0.52\times m\Rightarrow m=2$
2 moles of urea is present in 1000 g solvent.
number of moles of the solute $n_B=2$

mass of the solute $w_B=2\times 60=120g$
mass of the solvent $w_A=1000g$
mass of the solution $w_S=1120g/mL$
density of the solution $d_S=1.12g/mL$
volume of the solution $v_S=1000mL$

Molartiy $=\frac{\text{moles}}{\text{volume of the solution (L)}}$
$M=\frac{2}{1000}\times 1000=2M$

B) $\Delta T_f=K_{f}m$
Here,
$\Delta T_f$ is the depression in freezing point
$K_f$ is the molal depression freezing point constant
$=1.86\times 2=3.72$
Freezing point$T_f=0{}^{\circ }C-3.72{}^{\circ }C=-3.72{}^{\circ }C$

C) If the solution is cooled to $-1.86{}^{\circ }C$, ice will not form. This is because the freezing point of the solution is $-3.72{}^{\circ }C$.

d) Relative lowering of vapor pressure = mole fraction of the solute
number of moles of the solute = 2
number of moles of the solvent $=\frac{1000}{18}$

$\text{mole fraction}=\frac{2}{2+\frac{1000}{18}}=\frac{2\times 18}{1036}=\frac{9}{259}$