Animated Solution for Chemistry - Solutions: Elevation in the boiling point for $1$ molal solution of glucose is $2\text{ K}$. The depression in the freezing point for $2$ molal solution of glucose in the same solvent is $2\text{ K}$. The relation between $K_b$ and $K_f$ is

Unveiling the Connection Between Boiling and Freezing

Imagine you are standing in a laboratory with two beakers in front of you. Both contain glucose dissolved in the exact same solvent. The first beaker holds a $1\text{ molal}$ solution that is bubbling away, its boiling point elevated by exactly $2\text{ K}$. The second beaker holds a $2\text{ molal}$ solution that is freezing solid, its freezing point depressed by that very same amount, $2\text{ K}$.

This beautiful symmetry is not a coincidence; it is a puzzle waiting to be solved. Our goal is to find the hidden relationship between the ebullioscopic constant ($K_b$) and the cryoscopic constant ($K_f$) for this mysterious solvent.

To bridge the gap between these two physical phenomena, we must call upon our master tools for colligative properties.

The elevation in boiling point is governed by the equation:

Similarly, the depression in freezing point is dictated by:

Here, $i$ is the van't Hoff factor, which accounts for the number of particles a solute breaks into. Since glucose is a non-electrolyte and does not dissociate in solution, its van't Hoff factor is simply $i=1$.

Let's focus our attention on the first beaker. We know the elevation in boiling point $\Delta T_b$ is $2\text{ K}$, and the molality $m$ is $1\text{ molal}$.

Substituting these known values into our boiling point equation, we get:

This gives us the raw mathematical structure for our first condition. It tells us that for this specific solution, the constant $K_b$ is numerically equal to $2$.

Now, let's shift our gaze to the second beaker. The depression in freezing point $\Delta T_f$ is also $2\text{ K}$, but the molality $m$ here is $2\text{ molal}$.

Substituting these values into the freezing point equation, we find:

Notice how the equations are shaping up. We now have two distinct expressions that both equal the same numerical value.

Since both the elevation in boiling point and the depression in freezing point resulted in a $2\text{ K}$ change, we can directly equate the right-hand sides of our two equations.

This means:

And there we have it! A simple rearrangement gives us our final, elegant relationship:

$$K_b=2K_f$$

It is a beautifully straightforward result born from a solid understanding of the core formulas.

Before we wrap up, let's ponder a twist. What if we had used an electrolyte like sodium chloride ($\text{NaCl}$) instead of glucose?

Because $\text{NaCl}$ dissociates into two ions (${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$), the van't Hoff factor $i$ would change to $2$. This would double the effect on both the boiling and freezing points for the same molality. Always keep a sharp eye on the nature of the solute, as it fundamentally alters the landscape of colligative properties!