Question

A solution containing 62 g ethyl glycol in 250 g water is cooled to $-{{10}^{\circ }}C$. If ${{K}_{f}}$ of water is 1.86 K Kg/mol, the amount of water (in g) separated as ice is:
(A) 32
(B) 48
(C) 16
(D) 64

Answer 1

Hint: The cryoscopic constant gives us a relation between the molality and the depression in freezing point. You can use the relation to find the mass of water in the solution and then subtract it from the given mass to get the answer.

Complete step by step answer:
To solve this, firstly let us calculate the depression in freezing upon addition of the ethyl glycol.
We know that generally, the freezing point of water is zero degree Celsius i.e. 273K.
Here, the solution is cooled to $-{{10}^{\circ }}C$ i.e. 263K.
Therefore, depression in freezing point will be 273 – 263 K = 10 K.
Now, the relation between the difference in freezing point and molarity is given as:
\[\Delta {{T}_{f}}={{K}_{f}}\times m\]
Here, ${{K}_{f}}$ is also given to us as 1.86 K Kg/mol.
Therefore, we can write that:
\[10K=1.86K\text{ Kg mo}{{\text{l}}^{-1}}\times m\]
‘m’ is the molality. We know that molality is the moles of solute per kilogram of solvent.
Therefore, m = $\dfrac{no.of\text{ moles of solute }}{volume\text{ of solvent}(Kg)}$
And we know that number of moles of a substance = $\dfrac{weight}{molecular\text{ weight}}$
Therefore, we can write that:
\[{{T}_{f}}={{K}_{f}}\times \dfrac{Weigh{{t}_{solute}}\times 1000}{Weigh{{t}_{solvent}}\times Molecular\text{ weigh}{{\text{t}}_{solvent}}}\]
We know the molecular weight of ethyl glycol is 62 g/mol.
Therefore,
\[\begin{align}
& 10K=1.86K\text{ }Kg\text{ }mo{{l}^{-1}}\times \dfrac{62g\times 1000}{Weigh{{t}_{solvent}}\times 62g/mol} \\
& Or,Weigh{{t}_{solvent}}=186g \\
\end{align}\]
Therefore, the amount of water separated as ice = 250 g – 186 g = 64g.
Therefore, the correct answer is option (D) 64.

Note: As cryoscopic constant give us a relation between the molality and the depression in freezing point, similarly there is another constant named ebullioscopic constant which gives us a relation between the molality and the elevation in boiling point and it is denoted as${{K}_{b}}$. Through the process of ebullioscopy and cryoscopy, we can calculate the value of unknown molar mass through a known constant.