Question

The freezing point of one molal NaCl solution assuming NaCl to be 100% dissociated in water is:
[Molal depression constant is 1.86]
(a) $-{{2.72}^{o}}C$
(b) $-{{3.72}^{o}}C$
(c) ${{2.72}^{o}}C$
(d) ${{3.72}^{o}}C$

Answer 1

There will be a decrease in freezing point of a solvent when non volatile particles are added to it. This decrease in the freezing point is called depression in freezing point. If T is the freezing point of the pure solvent and ${{T}_{f}}$ is the freezing point of the solution, then the depression of freezing point $\Delta {{T}_{f}}=T-{{T}_{f}}$ .This idea has many applications.

Complete step by step solution: Depression in freezing point is a colligative property and is observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than the pure solvent and also it is directly proportional to molality of the solute.
The depression in freezing point is given by:
$\Delta {{T}_{f}}=i\times {{K}_{f}}\times m$
$\Delta {{T}_{f}}=T-{{T}_{f}}$
Here $\Delta {{T}_{f}}$= the change in boiling point = 0.81K
${{T}_{f}}$= freezing point of the solution
$T$= freezing point of the pure solvent
i= van’t Hoff factor
${{K}_{f}}$= Freezing point constant= 1.86 K/Kg mol
M= molality
And we know that freezing point of water =T= ${{0}^{o}}C$
The reaction is given as
$NaCl\to N{{a}^{+}}+C{{l}^{-}}$
It is given that NaCl dissociation is 100%. That means that it dissolved in water completely. So according to the above reaction two ions are formed: $N{{a}^{+}}$ and $C{{l}^{-}}$
So, i=2
Therefore, after substituting the values we get,
$\Delta {{T}_{f}}=2\times {{1.86}^{o}}C/kgmol\times 1m={{3.72}^{o}}C$
$\Delta {{T}_{f}}=T-{{T}_{f}}$=${{3.72}^{o}}C$
${{T}_{f}}$=$-{{3.72}^{o}}C$

Thus, the freezing point of 1 molal NaCl solution assuming NaCl to be 100% dissociated in water is$-{{3.72}^{o}}C$. The correct option is (b).

Note: Whenever we calculate decrease in freezing point$\Delta {{T}_{f}}$, it is the difference between the freezing point of pure solvent and freezing point of the solution. Not the other way round.