Question

Given,31g of ethylene glycol(${{C}_{2}}{{H}_{6}}{{O}_{2}}$) is mixed with 500g of solvent(${{K}_{f}}$ of the solvent is $2K\,kg\,mo{{l}^{-1}}$). What is the freezing point of the solution(in K)?
A. 272
B.271
C.270
D.274
E. 275

Answer 1

Hint: Substitute the data given in the question in the formula for the difference in freezing point, and then subtract it from the normal freezing point which is 273K, to find the new one.

Complete answer:
In order to answer our question, we need to learn about the depression in freezing point. Now, we are all aware of Raoult’s law. Suppose a solvent is taken and no other non volatile solute is added to it. So, the solvent has a specific boiling point and a freezing point. Now, suppose a non volatile solute for example, common salt is added to the solvent, which is water. Now, the non volatile solute occupies some space in between the solvent molecules. So, these non volatile solutes contain some heat energy. In order to freeze the entire solution, the heat from both the solute as well as the solvent needs to be removed. Presence of a non volatile solute means that more heat needs to be extracted from the solution, in order to freeze it completely. That is why, when a non volatile solute is added to the solution, a depression in freezing point is observed. Now, the formula for the difference in freezing point is given by the formula:
\[\Delta {{T}_{f}}=i\times {{K}_{f}}\times m\]
Here, $\Delta {{T}_{f}}$ is the difference in freezing point, I is the Van’t Hoff factor and m is the molality of the solution. In the case of ethylene glycol, the Van’t Hoff factor is equal to 1, as it is not an ionic compound. The molar mass of ethylene glycol is 62. Now, we will substitute the data given in the question to find the difference in freezing point:
\[\begin{align}
  & \Delta {{T}_{f}}={{K}_{f}}\times \left( \dfrac{{{W}_{B}}}{{{M}_{B}}} \right)\times \left( \dfrac{1000}{{{W}_{A}}} \right) \\
 & \Rightarrow \Delta {{T}_{f}}=\dfrac{2\times 31\times 1000}{62\times 500}=2K \\
\end{align}\]

So, the new freezing point comes out to be $(273-2)K=271K$, which gives us option B as the correct answer.

Note:
It is to be noted that ${{K}_{f}}$ is known as the freezing point constant and it gives us an idea of how much the solution can freeze. More the amount of solute added to the solution, more will be the depression in the freezing point.