Question

The molal freezing point constant of water is ${{1.86}^{\circ }}C/m$ . Therefore, the freezing point of 0.1M NaCl solution in water is expected to be?
[A] $-{{1.86}^{\circ }}C$
[B] $-{{0.186}^{\circ }}C$
[C] $-{{0.372}^{\circ }}C$
[D] $+{{0.372}^{\circ }}C$

Answer 1

To solve this you must know that the molal freezing constant is also known as the cryoscopic constant. Here, we can use the relation between depression in freezing point and cryoscopic constant to find the required freezing point. The relation is: $\Delta {{T}_{f}}={{K}_{f}}\times m$.

Complete step by step answer:
Firstly, we should know that the molal freezing point constant is nothing but the cryoscopic constant i.e. ${{K}_{f}}$.
The cryoscopic constant gives us a relation between the molality and the depression in freezing point that we can use here. The relation is -
$\Delta {{T}_{f}}={{K}_{f}}\times m$

Here, $\Delta {{T}_{f}}$ is the depression in freezing point and ‘m’ is the molality.
In the question, m is given to us as 0.1 and ${{K}_{f}}$ is also given as ${{1.86}^{\circ }}C/m$.
Also, we know that the freezing point of water is 0 degree Celsius. So, let us assume that the freezing point of NaCl is x.
Therefore, from the above relation we can write that:
$\left( 0-x \right)=1.86\times 0.1={{0.186}^{\circ }}C$

However, we must remember that sodium chloride ionises as $N{{a}^{+\text{ }}} and \text{ C}{{\text{l}}^{-}}$ ions. So, whatever depression we calculate from here the actual value will be twice of that. This is also known as the Van’t Hoff factor i.e. the number of splitting or forming of ions of the solute in a solution.
Therefore, actual freezing point will be $2\times -{{0.186}^{\circ }}C=-{{0.372}^{\circ }}C$
So, the correct answer is “Option C”.

Note: Here, the freezing point changed due to addition of NaCl. This lowering in freezing point is a colligative property of a liquid. The property of solution that changes upon addition of solute is known as the colligative properties.
There are 4 colligative properties shown by liquids. They are-osmotic pressure, depression of freezing point, elevation of boiling point and relative lowering of vapour pressure.