Question

45g of ethylene glycol $\left( {{{\text{C}}_2}{{\text{H}}_6}{{\text{O}}_2}} \right)$ is mixed with $600\;{\text{g}}$ of water. The freezing point of the solution is $\left( {{{\text{K}}_{\text{f}}}} \right.$ for water is $\left. {1.86\;{\text{K}}\;{\text{kg}}\;{\text{mo}}{{\text{l}}^{ - 1}}} \right)$
(A) $273.95\;{\text{K}}$
(B) $270.75\;{\text{K}}$
(C) $370.95\;{\text{K}}$
(D) $373.95\;{\text{K}}$

Answer 1

If any substance or solute is added to any solvent, the decrease occurs at its freezing point. In fact, the decrease in the freezing point of a solvent on the addition of a non-volatile solute is a colligative property called depression in the freezing point.

Formula used:
The freezing point depression can be expressed as
$\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}} \times {\text{m}} \times {\text{i}}$
Where
$\Delta {{\text{T}}_{\text{f}}}$ is the freezing-point depression expressed as ${{\text{T}}_{\text{f}}}$ (pure solvent) $- {{\text{T}}_{\text{f}}}($ solution $).$
${{\text{K}}_{\text{f}}}$ is a cryoscopic constant which only depends on the properties of the solvent.
${\text{m}}$ is the molality which is represented as moles solute per kilogram of solvent
${\text{i}}$ is the van't Hoff factor.

Complete step by step solution:
Now, we will have to find the molarity by finding the number of moles of ${{\text{C}}_2}{{\text{H}}_6}{{\text{O}}_2}$ (ethylene glycol)
Number of moles can be calculated using the formula
moles $= \dfrac{{{\text{ mass of substance }}}}{{{\text{ molar mass }}}}$
Molar mass of ethylene glycol ${{\text{C}}_2}{{\text{H}}_6}{{\text{O}}_2}$ $= 62{\text{g}}$
The following information is provided to us in the question:
Mass of water $= 600{\text{g}}$
Mass of ethylene glycol $= 45{\text{g}}$
${{\text{K}}_{\text{f}}} = 1.86\;{\text{Kkgmo}}{{\text{l}}^{ - 1}}$
Now, let us continue finding the molality,
That is, $\dfrac{{45 \times 1000}}{{62 \times 600}} = 1.209$
We had to divide the molality by $1000$ because we had to calculate in terms of kg
Now, we will put all the known values in our formula to get the required answer
$\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}} \times {\text{m}} \times {\text{i}}$
$\Rightarrow \Delta {{\text{T}}_{\text{f}}} = 1.86 \times 1.209 \times 1$
Upon further solving, we get
$\Delta {{\text{T}}_{\text{f}}} = 2.25 K$
The freezing point of the solution is given by the formula
${T_{f(pure solvent)}} - {T_{f(solution)}}$
Now, let us substitute the values of
${T_{f(pure solvent)}} = 273.2 K$ and $\Delta {T_{f(solution)}} = 2.25 K$ in the above formula to get the final required answer
So, the freezing point of the solution is given by
$273.2 K - 2.25 K = 270.95 K$
Hence, the correct option is (A.)

Note:
The common mistake committed by learners here is not to correctly use the molality formula. As the formula is moles of solute per kg of solvent, do not forget to divide by $1000$ . So, remember to convert the solvent mass that can be given in grams into kilograms to convert that. Depression is a colligative property at the freezing point that depends on the number of particles in the solution. The greater the number of particles, the stronger the colligative property will be.