Animated Solution for Chemistry - Solutions: For an aqueous solution, freezing point is $-0.186^{\circ }\text{C}$. Elevation of the boiling point of the same solution is ($K_f=1.86^{\circ }{\text{ mol}}^{-1}\text{ kg}$ and $K_b=0.512^{\circ }{\text{ mol}}^{-1}\text{ kg}$)

Imagine you have a beaker filled with an aqueous solution. We don't know what the solute is, and we don't know how much of it is dissolved. But we do know one crucial piece of information: the freezing point of this solution has dropped to $-0.186^{\circ }\mathrm{C}$.

Since it is an aqueous solution, the solvent is water, which normally freezes at exactly $0^{\circ }\mathrm{C}$. This means the presence of the solute has caused a depression in the freezing point ($\Delta T_f$).

We can easily calculate this depression:

We are also given the molal depression constant ($K_f$) and the molal elevation constant ($K_b$) for water. Our goal is to find the elevation in the boiling point ($\Delta T_b$) for this exact same solution.

Here is the beautiful part about colligative properties: they depend purely on the number of solute particles, which is quantified by the molality ($m$) of the solution.

Let's write down the formulas for both phenomena:

Because we are dealing with the exact same solution in both cases, the molality ($m$) is identical. This molality acts as a mathematical bridge connecting the freezing point depression to the boiling point elevation.

Instead of calculating the molality first and then plugging it into the second equation, we can be smart and simply take the ratio of the two equations:

The molality ($m$) beautifully cancels out, leaving us with a direct, elegant relationship:

Now, it is just a matter of substituting the values we know into our derived relationship.

We rearrange the formula to solve for $\Delta T_b$:

Substitute the given values:

Look closely at the numbers; they are designed to be friendly! Notice how $0.186$ and $1.86$ are related. Dividing $0.186$ by $1.86$ gives exactly $0.1$.

And there we have it! The elevation in the boiling point of this solution is $0.0512^{\circ }\mathrm{C}$.

What if the solute dissolved in the water was an electrolyte, like sodium chloride ($\text{NaCl}$)? Would our approach change?

If it were an electrolyte, it would dissociate into ions, and we would have to introduce the van't Hoff factor ($i$) into our equations:

But watch what happens when we take the ratio again:

Both the van't Hoff factor ($i$) and the molality ($m$) cancel out! This proves that our shortcut ratio formula is universally applicable to any solute, whether it is a non-electrolyte like glucose or a strong electrolyte like salt, as long as we are comparing the same solution.