Animated Solution for Chemistry - Solutions: $C_6{H}_6$ freezes at $5.5^{\circ }C$. The temperature at which a solution $10\text{ g}$ of $C_4{H}_{10}$ in $200\text{ g}$ of $C_6{H}_6$ freeze is .........${}^{\circ }C$. (The molal freezing point depression constant of $C_6{H}_6$ is $5.12^{\circ }C/m$.)

Imagine you have a beaker filled with pure benzene ($C_6{H}_6$). Normally, it freezes at a crisp $5.5^{\circ }C$. But now, we are going to add $10\text{ g}$ of butane ($C_4{H}_{10}$) into $200\text{ g}$ of this benzene.

What happens? The freezing point drops! This phenomenon is known as the depression in freezing point, which is a colligative property. It depends purely on the number of solute particles added to the solvent, not on their identity.

To figure out exactly how much the freezing point drops, we need our trusty formula for the depression in freezing point:

Here, $i$ is the van't Hoff factor, $K_f$ is the molal freezing point depression constant, and $m$ is the molality of the solution. Now, since butane is a covalent organic compound and doesn't break apart into ions in the solution, its van't Hoff factor, $i$, is simply $1$.

Let's break down the molality part. Molality is the moles of our solute (butane) divided by the mass of our solvent (benzene) in kilograms. The molar mass of butane is $12\times 4+10$, which is $58\text{ g/mol}$. Let's plug all our known values into the master equation:

Alright, let's crunch the numbers. $200$ divided by $1000$ is $1/5$. When we flip that fraction up, we get $5.12\times 50$, all divided by $58$.

If you calculate this carefully, you will find that the temperature drop, $\Delta T_f$, is approximately $4.414^{\circ }C$.

We are almost there! We know the pure benzene freezes at $5.5^{\circ }C$. The new freezing point ($T_f^{\prime }$) is simply the original freezing point minus the drop we just calculated.

Since the question asks for the nearest integer, our final answer is $1^{\circ }C$! Always pay close attention to whether your solute dissociates or associates in the given solvent. If it were an electrolyte, the van't Hoff factor would be greater than one, causing a much larger drop in the freezing point. That's a classic trap!