Question

The freezing point of a solution containing $50\;c{m^3}$ of ethylene glycol in $50g$ of water is found to be $- {34^o}C$ . Assuming dilute solution, the density of solution is: ( ${K_f}$ for ${H_2}O = 1.86\;{\text{Kmolalit}}{{\text{y}}^{ - 1}}$ )
A. $1.133\;gc{m^{ - 3}}$
B. $2.133\;gc{m^{ - 3}}$
C. $0.133\;gc{m^{ - 3}}$
D. $1.62\;gc{m^{ - 3}}$

Answer 1

Hint : Properties of solutions which do not depend on the identity of solute but on the concentration of solute molecules or ions are known as colligative properties of the solutions. These properties include lowering in vapor pressure, elevation in boiling point, depression in freezing point and osmotic pressure.

Complete Step By Step Answer:
According to question, the given data is as follows:
Volume of ethylene glycol (solute) $= 50\;c{m^3}$
So, the mass of ethylene glycol ${M_{eg}} = 50 \times d$ , where d is the density of solution.
Mass of water (solvent) $= 50\;g$
$\Rightarrow {M_{{H_2}O}} = 0.05\;kg$
Final temperature $T_f^s = - {34^o}C$
So, the value of depression in freezing point $\Delta {T_f} = T_f^o - T_f^s$
$\Rightarrow \Delta {T_f} = 0 - ( - 34)$
$\Rightarrow \Delta {T_f} = {34^o}C$
With these known values, we can calculate the molality of the solution as follows:
molality $= \dfrac{{{\text{number of moles of solute }}}}{{{\text{mass of solvent in kg}}}}$
Substituting values:
$\Rightarrow m = \dfrac{{d \times 50}}{{62 \times 0.05}}$
$\Rightarrow m = 16.13d$
We know that depression in freezing point can be expressed in terms of cryoscopic constant and molality as per following expression:
$\Delta {T_f} = {K_f}m$
Where, ${K_f}$ is the cryoscopic constant and m is the molality of the solution.
Substituting values in the expression, then the density of the solution can be calculated as follows:
$\Rightarrow 34 = 1.86 \times 16.13d$
$\Rightarrow d = 1.133\;gc{m^{ - 3}}$
Hence, the density of the solution is $1.133\;gc{m^{ - 3}}$ . So, option (A) is the correct answer.

Note :
It is important to note that the addition of a non-volatile solute leads to decrease in the vapour pressure of the solution which causes the shift of equilibrium of solid and liquid phase at lower temperature and thus, increasing the molality of the solute leads to depression in the freezing point of solvent.