Question

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to $4{\text{ kg}}$ of water to prevent it from freezing at $- 6^\circ C$ will be (${K_f}{\text{ of water = 1}}{\text{.86K}}\,{\text{kg mo}}{{\text{l}}^{ - 1}}$) and molar mass of ethylene glycol$= 62g{\text{ mo}}{{\text{l}}^{ - 1}}$)
(A) $804.32{\text{gm}}$
(B) $204.30{\text{gm}}$
(C) $400.00{\text{gm}}$
(D) $304.60{\text{gm}}$

Answer 1

Depression in freezing point is a colligative property observed in solutions with non-volatile solute. The freezing points of solutions are lower than that of the pure solvent and is directly proportional to molality of the solution.

Complete answer: or Complete step by step answer:
The decrease in the freezing point of the solute is known as freezing-point depression which can be represented as $\Delta {T_f}$
To proceed for the calculation of the mass of ethylene glycol the following data has been given below:
Weight of solvent$\left( {{H_2}O} \right) = 4kg = 4000g$
Freezing temperature of solution $\left( {{T_f}} \right) = - 6^\circ C$
Molal depression constant $\left( {{K_f}} \right) = 1.86{\text{ }}Kkgmo{l^{ - 1}}$
Molar mass of ethylene glycol$= 62g/mol$
Mass of ethylene glycol can be calculated by using cryoscopy. We know that depression in freezing (cryoscopy) point is given by,
$\Delta {T_f} = {K_f} \times m{\text{ }}.........{\text{ }}\left( i \right)$
Where $m$ is the molality of the solution.
Molality is defined as the moles of solute per kilogram of the solvent. It is the property of the solution.
Molality is further given by,

$$molality\left( m \right) = \dfrac{{{\text{number of moles of solute}}\left( n \right)}}{{{\text{mass of solvent }}\left( {inKg} \right)}}{\text{ }}\left( {{\text{number of moles}} = \dfrac{{{\text{mass of substance}}}}{{{\text{molecular mass of substance}}}}} \right) \\\ molality\left( m \right) = \dfrac{{{\text{mass of solute}}\left( w \right) \times 1000}}{{{\text{molecular mass of solute }}\left( M \right) \times {\text{mass of solvent}}\left( {in{\text{ gm}}} \right)}} \\\$$

Putting the value of molality in equation (i),
$\Delta {T_f} = {K_f} \times \dfrac{w}{{MW}} \times 1000$………….. (ii)
Where, $\Delta {T_f}$is depression in freezing point.
${K_f}$ is the molal depression constant and it is the proportionality constant (it is also called cryoscopic constant).
$\begin{gathered} {\text{w}} = {\text{mass of solute}} \\\ {\text{M}} = {\text{molecular mass of solute}} \\\ \end{gathered}$
$W$ Is the weight of solvent$\left( {{H_2}O} \right)$. Also,
$\therefore \Delta {T_f} = {T_f}^\circ - {T_f}{\text{ }}\left( {where{\text{ }}{{\text{T}}_f}^0{\text{ }}\left( {0^\circ C} \right){\text{ is the freezing point of pure water}}} \right)$.
$= 0 - \left( { - 6} \right)$
$= 6^\circ C$
Now Form equation (ii),
$\Delta {T_f} = {K_f} \times \dfrac{w}{{M \times W}} \times 1000$
Putting the value of depression in freezing point, molal depression constant, weight of solute, molecular weight of solute and weight of solvent, we get
$6 = 1.86 \times \dfrac{w}{{62 \times 4000}} \times 1000$
$\therefore w = \dfrac{{6 \times 62 \times 4000}}{{1.86 \times 1000}} = 800gm.$

So, the correct answer is “Option A”.

Note:
It must be clear to the students that the solute will be non-volatile. Calculations must be done carefully and if the question asks about freezing point of solution then the reference is taken to be water. Water is taken as a universal solvent and its freezing point is$0^\circ C$. Cryoscopy is used to find out the molar mass of solute as well.