Question

Which of the following has a lower freezing point?
A. $0.1m$ sucrose
B. $0.1m$ urea
C. $0.1m$ common salt
D. $0.1m$ glucose

Answer 1

We know that the freezing point of the solution depends upon the concentration of the solute and well the nature of the solute whether it is ionic or covalent. For the given question, we should know the factors that are affecting the freezing point of the solution.

Complete answer:
Let us first understand the very basic concepts related to solutions. We have read many times about the colligative properties of the solutions. So, depression in freezing point is one of the colligative properties of the solutions.
We know the formula of this colligative property i.e. depression in freezing point. By looking at the formula we come to know that the freezing point of the solution depends upon the Van't Hoff factor and the concentration of the solutions.
The formula is as follows:
$\Delta {T_f} = i \times {K_f} \times m$
Where, $\Delta {T_f}$ denotes the change in freezing point, $i$denotes the Van’t Hoff Factor, ${K_f}$is the cryogenic constant and the last one i.e. $m$indicates the molality of the solution.
Now, in the given question, we will compare the concentration values and the Van’t Hoff factors of all the given solutions.
A. $0.1m$ sucrose
Here, sucrose is the non-ionic compound, so the value of $i$ is $1$.
$\Delta {T_f} = 1 \times {K_f} \times 0.1$
$\Delta {T_f} = 0.1{K_f}$
B. $0.1m$ urea
In this case also, the value of $i$ is $1$ for urea.
$\Delta {T_f} = 1 \times {K_f} \times 0.1$
$\Delta {T_f} = 0.1{K_f}$
C. $0.1m$common salt
Here, common salt i.e. $NaCl$. It is an ionic compound, so the value of $i$ is $2$.
$\Delta {T_f} = 2 \times {K_f} \times 0.1$
$\Delta {T_f} = 0.2{K_f}$
D. $0.1m$glucose
Here, the value of $i$ is $1$ in case of glucose.
$\Delta {T_f} = 1 \times {K_f} \times 0.1$
$\Delta {T_f} = 0.1{K_f}$
Thus, we conclude that the Van’t Hoff factor of common salt i.e. $NaCl$ is $2$. Therefore, it produces most ions and it lowers the freezing point to the maximum value.
The correct answer is option (C).

Note:
We observed that the change in freezing point depends upon the concentration of the solute present in the given solution. It mostly depends upon the number of the solute rather the kind of the particles present. These properties are called colligative properties.