Question

Which of the following solutions has maximum freezing point?
A. $1$ molar of $NaCl$ solution
B. $1$ molar of $KCl$ solution
C. $1$ molar of $CaC{l_2}$ solution
D. $1$ molar of urea solution

Answer 1

Hint: We know that the colligative properties depend on the number of particles.
In solutions, we have some colligative properties whose value depends on the number of particles present in the system. These include elevation in the boiling point for solvent, relative lowering of the vapor pressure for solvent, depression in the freezing point for solvent and the osmotic pressure for the solution.

Complete step by step answer:
Let’s have a brief look at the depression in freezing point of a solvent which is volatile when a non-volatile solute is added to it. We know that states of matter can be inter-converted. When matter is converted from its liquid state to solid state, the temperature at which both the phases are in equilibrium with each other is called freezing point. We can also define it in terms of vapor pressure as the temperature at which the vapor pressure in liquid state is equal to that in solid state.
We know that addition of a non-volatile solute to a volatile solvent decreases the vapor pressure of the solvent which in turn lower the freezing point as well. We can represent it with the following diagram:

We know that the freezing point of pure solvent $\left( {T_f^0} \right)$ and that of the solution $\left( {{T_f}} \right)$are related to the depression in the freezing point $\left( {\Delta {T_f}} \right)$ as follows:
$\Delta {T_f} = T_f^0 - {T_f}$
We also have a relationship between $\Delta {T_f}$and the molality $\left( m \right)$ of the solution as follows:
$\Delta {T_f} = {K_f}m$
Here, ${K_f}$ is the molal depression constant for a given solvent.
As colligative properties are a function of number of particles, we also need to have a look at the Van’t Hoff factor $\left( i \right)$ which can be expressed as follows:
$i = \dfrac{{{\rm{Total number of moles of particles finally present in the solution}}}}{{{\rm{Number of moles of particles initially taken}}}}$
After incorporating Van’t Hoff factor, we get:
$\Delta {T_f} = i{K_f}m$
Now let’s have a look at the given compounds and their quantities. The solvent is common in all cases so ${K_f}$ is the same. Now, we have to calculate the molality $\left( m \right)$ from the molarity $\left( M \right)$ which can be done as follows:
$\begin{array}{c}
M = \dfrac{{{\rm{number of moles}}}}{{{\rm{1 L solution}}}}\\
{\rm{number of moles}} = M \times {\rm{1 L solution}}
\end{array}$
Substituting this for molality:
$m = \dfrac{{M \times {\rm{1 L solution}}}}{{{\rm{1 kg solvent}}}}$
Here, we are assuming that the densities of all the solutions are equal and we can simply write:
$m = M$
So now as given, molality of all the solutions are also equal which makes Van’t Hoff factor $\left( i \right)$ to be the deciding factor.
Let’s calculate the value of Van’t Hoff factor for each solute as follows:
$\begin{array}{c}
NaCl \to N{a^ + } + C{l^ - }\\
KCl \to {K^ + } + C{l^ - }\\
CaC{l_2} \to C{a^{2 + }} + 2C{l^ - }
\end{array}$
As we can see that the electrolytic salts get dissociated into the solution and have $i = 2,2\;{\rm{and}}\;3$ respectively. However, ammonia remains as it is with $i = 1$. So, depression in the freezing point would be least for ammonia and its solution would have the maximum freezing point.

Therefore, from the above explanation the correct option is (D).

Note:
We have to be careful with the effect on the value of depression in freezing point and that of freezing point as minimum depression would mean the freezing point of the solution is closer to the that of the pure solvent.