Question

Given that the freezing point depression constant for water is $1.86\,C\,kg\,mo{{l}^{-1}}$. How do you calculate the change in freezing point for a 0.833m sugar solution?

Answer 1

Hint: The values given in the question need to be substituted in the formula for finding out the depression in freezing point, in order to obtain the answer.

Complete answer:
In order to answer the question, we need to learn about the depression in freezing point. When a non volatile solute is added to a solvent, then the solute forms a layer above the solvent. Freezing point of a substance is the temperature at which solid and liquid phases of the substance coexist. It is defined as the temperature at which its liquid and solid phases have the same vapour pressure.
The freezing point of a pure liquid remains constant. Now, in case a non-volatile solute is dissolved in the pure liquid to contain a solution, the freezing point gets lowered. The freezing point of solution refers to the temperature at which the vapour pressure of the solvent in two phases, i.e., liquid solution and solid solvent is the same. Since, the vapour pressure of solvent in solution is reduced, it becomes equal to that of the solid solvent if temperature is lowered.
Now, let us come to the question. If m is the molality and ${{k}_{f}}$ is the freezing point depression constant, then the depression in freezing point is calculated by the formula:
\[\Delta {{T}_{f}}={{k}_{f}}\times m\]
Now, by substituting the values in the above formula, we get the depression in freezing point as:
\[\begin{align}
  & \Delta {{T}_{f}}=1.86K\,kg\,mo{{l}^{-1}}\times 0.833\,mol\,k{{g}^{-1}} \\
 & \Delta {{T}_{f}}=+1.55K \\
\end{align}\]
This is the required change in freezing point for a 0.833m sugar solution.

Note:
It is to be noted that the actual formula for calculating the depression in freezing point is $\Delta {{T}_{f}}=i\times {{k}_{f}}\times m$, but as sugar is a nonelectrolyte, we take the value of i as 1, where i is the Van’t Hoff factor.