Question

A solution of 1.25g non-electrolyte in 20g water freezes at 271.94K. If ${{K}_{f}}$ is 1.86 K Kg/mol then molecular mass of solute will be-
[A] 207.8
[B] 179.79
[C] 209.6
[D] 109.65

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 molality and then from there you can find the molecular mass.

COMPLETE STEP BY STEP SOLUTION: To solve this, firstly let us calculate the depression in freezing upon addition of the electrolyte.
We know that the freezing point is 0 degree Celsius i.e. 273K.
Upon addition of the non-electrolyte, the freezing point changes to 271.94K.
Therefore, depression in freezing point = 273 – 271.94 K = 1.06K.
The relation between the difference in freezing point and molarity is given as:
\[\Delta {{T}_{f}}={{K}_{f}}\times m\]
‘m’ is the molality. We know that molality is the moles of solute per kilogram of solvent.
Here, the solvent is water and it’s 20g i.e. 0.02Kg.
Therefore, we can write that:
\[\begin{align}
 & \Delta {{T}_{f}}={{K}_{f}}\times \dfrac{number\text{ of moles}}{0.02} \\
& or,number\text{ of moles =}\dfrac{\text{1}\text{.06 K }\times \text{0}\text{.02Kg}}{1.86\text{ K Kg mol}{{\text{l}}^{-1}}}\text{ = 0}\text{.01139}\simeq \text{0}\text{.0114mol } \\
\end{align}\]
Now, we know that number of moles of a substance = $\dfrac{weight}{molecular\text{ weight}}$
Therefore, molecular weight = \[\dfrac{Weight}{number\text{ of moles }}~\]
Putting the values we will get, molecular weight = \[~\dfrac{1.25g}{0.0114mol}~\] = 109.65 mol.

Therefore, the correct answer is option [D] 109.65.

Now, let us assume that the mass of the non-electrolyte is ${{W}_{2}}$ and it is given to us as 1.25g.
And let the mass of the solvent is ${{W}_{1}}$ and it is given to us as 20g.

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.