Animated Solution for Chemistry - Solutions: 40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is ............ K. (Nearest integer) [Given, $K_f=1.86{\text{ K kg mol}}^{-1}$, density of water $=1.00{\text{ g cm}}^{-3}$, freezing point of water $=273.15\text{ K}$]

Welcome to the fascinating and deeply intuitive world of colligative properties! Imagine you are standing in a freezing, snow-covered landscape, holding a beaker of pure, crystal-clear water. You know from basic science that at exactly $273.15\text{ K}$ (or $0^{\circ }\text{C}$), this water will begin to turn into solid ice. The water molecules will slow down, lose their kinetic energy, and arrange themselves into a beautiful, highly ordered hexagonal crystal lattice.

But what happens when you introduce an impurity into this pristine environment? What happens when you dissolve $40\text{ g}$ of glucose into this water?

In our beaker, we have $200\text{ mL}$ of pure water. This water acts as our solvent—the medium in which everything else will happen. To this solvent, we add $40\text{ g}$ of glucose, chemically known as $C_6{H}_{12}{O}_6$.

Glucose is a non-volatile, non-electrolyte solute. This means it doesn't evaporate easily, and more importantly, it doesn't break apart into smaller ions when it dissolves. It stays intact as whole glucose molecules. When these bulky glucose molecules dissolve, they scatter throughout the water, disrupting the orderly arrangement of the water molecules.

Because of this disruption, the water molecules find it much harder to pack together and form that solid crystal lattice. They need to be slowed down even more. In other words, the water needs to be cooled to an even lower temperature to freeze. This beautiful and universal phenomenon is termed as the depression in freezing point.

To quantify this exact drop in temperature, we don't just guess; we rely on a simple yet incredibly powerful mathematical relationship. The depression in freezing point, denoted by the symbol $\Delta T_f$, is directly proportional to the concentration of the solute particles in the solution. Specifically, it is proportional to the molality.

The master equation that governs this behavior is:

Let's break down the anatomy of this equation. Here, $K_f$ is the molal depression constant, also known as the cryoscopic constant. It is a unique, intrinsic property of the solvent itself. It tells us how much the freezing point will drop for every $1\text{ molal}$ increase in concentration. For water, $K_f$ is given as $1.86{\text{ K kg mol}}^{-1}$.

The variable $m$ represents the molality of the solution. Notice that we use molality, not molarity. Why? Because molality is based on the mass of the solvent, which remains completely unaffected by changes in temperature. Molarity, on the other hand, is based on volume, which expands or contracts as the temperature drops.

Molality is defined as the number of moles of solute dissolved per kilogram of solvent. Let's calculate this step by step.

First, we need to find the exact number of moles of our solute, glucose. We are given $40\text{ g}$ of glucose. The molar mass of glucose is $180{\text{ g mol}}^{-1}$.

So, we have approximately $0.222$ moles of glucose swimming around in our beaker.

Next, we need the mass of our solvent in kilograms. We are given $200\text{ mL}$ of water. How do we convert this volume into mass? We use the density! The density of water is exactly $1.00{\text{ g cm}}^{-3}$ (which is the same as $1.00{\text{ g mL}}^{-1}$). Therefore, the mass of the water is exactly $200\text{ g}$.

To convert this to kilograms, we divide by $1000$:

Now, we have everything we need to find the molality. We divide the moles of solute by the mass of the solvent in kilograms:

Our solution has a concentration of $1.11\text{ molal}$.

With the molality firmly in hand, we can now calculate exactly how much the freezing point will drop. We substitute our values back into the master equation:

This is a profound result. It means the freezing point of the water has been pushed down by $2.066\text{ K}$ simply because we added some glucose.

To find the new, actual freezing point of the solution, we must subtract this depression from the original freezing point of pure water. Pure water freezes at $273.15\text{ K}$.

The question specifically asks for the nearest integer. When we round off $271.084$, we arrive at our final, elegant answer: $271\text{ K}$.

This problem perfectly illustrates the power and predictability of colligative properties. But as elite students, we must always push our intuition further. What if we changed the rules of the game? What if, instead of glucose, we had dissolved $40\text{ g}$ of sodium chloride ($\text{NaCl}$)?

Unlike glucose, sodium chloride is a strong electrolyte. When it hits the water, it doesn't stay intact. It violently dissociates into ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions. This means that for every one mole of $\text{NaCl}$ you dissolve, you actually get two moles of particles swimming in the solution.

Because colligative properties depend strictly on the number of particles and not their identity, this introduces a crucial new variable into our master equation: the van't Hoff factor, denoted by $i$.

For $\text{NaCl}$, the van't Hoff factor $i$ is approximately $2$. This means the freezing point would drop almost twice as much as it did with glucose!

Always pay close attention to the chemical nature of your solute. Is it an electrolyte or a non-electrolyte? Does it associate or dissociate? Asking yourself these questions will completely change the landscape of your solution and ensure you never fall into a trap. Keep this powerful concept in mind as you tackle more complex challenges!