Animated Solution for Chemistry - Solutions: 1 g of a non-volatile, non-electrolyte solute is dissolved in 100 g of two different solvents A and B, whose ebullisocopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, $\frac{\Delta T_b(A)}{\Delta T_b(B)}$, is

The phenomenon of boiling point elevation is one of the most fascinating consequences of colligative properties in physical chemistry. When you dissolve a non-volatile solute into a pure solvent, you fundamentally alter the physical behavior of that solvent. The vapor pressure drops, and as a direct result, the temperature required to make the liquid boil increases. But how do we quantify this change? And what happens when we compare two completely different solvents under identical conditions?

This problem from the JEE Main 2019 paper is a beautiful exercise in conceptual clarity. It strips away the need for heavy calculations and instead tests your fundamental understanding of the mathematical relationships governing colligative properties. Let's embark on a detailed journey to unravel the elegance of this problem.

Imagine you are standing in a laboratory with two identical beakers in front of you. In the first beaker, you carefully pour exactly $100\text{ g}$ of Solvent A. In the second beaker, you pour $100\text{ g}$ of a completely different liquid, Solvent B.

Now, you take a non-volatile, non-electrolyte solute. Think of a simple sugar like glucose or sucrose. You weigh out exactly $1\text{ g}$ of this solute and dissolve it into Solvent A. You then weigh out another $1\text{ g}$ of the exact same solute and dissolve it into Solvent B.

The problem provides us with one crucial piece of comparative data: the ratio of their ebullioscopic constants. The ebullioscopic constant, denoted as $K_b$, is a unique property of every solvent. It tells us how much the boiling point of that specific solvent will elevate for a $1\text{ molal}$ concentration of solute. We are given that:

Our ultimate goal is to find the ratio of the elevation in their boiling points, which is mathematically represented as $\frac{\Delta T_b(A)}{\Delta T_b(B)}$.

To solve any problem involving boiling point elevation, we must rely on the master equation of colligative properties:

Let's break down the anatomy of this equation: - $\Delta T_b$ is the elevation in boiling point. - $i$ is the van't Hoff factor, which accounts for the number of particles the solute splits into upon dissolution. - $K_b$ is the ebullioscopic constant of the solvent. - $m$ is the molality of the solution.

The beauty of this problem lies in recognizing which of these variables are changing and which are remaining perfectly constant across our two beakers.

Let's take a closer look at the molality term, $m$. Molality is a measure of concentration defined as the number of moles of solute per kilogram of solvent. The formula for molality is:

Where: - $w_2$ is the mass of the solute. - $M_2$ is the molar mass of the solute. - $w_1$ is the mass of the solvent.

Now, let's apply this to our specific laboratory setup. For both Beaker A and Beaker B, we added exactly $1\text{ g}$ of the solute. Therefore, $w_2=1\text{ g}$ for both. Because we used the exact same solute in both beakers, the molar mass $M_2$ is identical for both. Finally, we used exactly $100\text{ g}$ of solvent in both beakers, meaning $w_1=100\text{ g}$ for both.

Since every single variable in the molality equation ($w_2$, $M_2$, and $w_1$) is identical for both solutions, we can confidently conclude that the molality of Solution A is perfectly equal to the molality of Solution B:

The concentration of the solute particles in both beakers is exactly the same!

Next, we must consider the van't Hoff factor, $i$. The problem explicitly states that our solute is a "non-electrolyte". An electrolyte, like sodium chloride ($\text{NaCl}$), would dissociate into multiple ions (${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$) when dissolved in a solvent, effectively multiplying the number of particles and increasing the colligative effect.

However, a non-electrolyte does not dissociate. When you dissolve one molecule of a non-electrolyte, it remains as one intact molecule in the solution. Therefore, the number of particles does not multiply.

For any non-electrolyte solute, the van't Hoff factor is simply:

Since we used the same non-electrolyte solute in both solvents, $i=1$ for both Solution A and Solution B.

Let's return to our master equation and substitute what we have discovered.

We have established that the molality $m$ is a constant value for both solutions. We also know that the van't Hoff factor $i$ is $1$ for both.

When you have an equation where multiple terms are constant, the relationship between the remaining variables becomes a direct proportionality. In this case, the elevation in boiling point depends entirely and exclusively on the ebullioscopic constant of the solvent:

This is the conceptual turning point of the problem. It tells us that whichever solvent has a higher $K_b$ value will experience a proportionally higher elevation in its boiling point, given that the concentration of particles is identical.

Because of the direct proportionality we just established, we can write the ratio of the boiling point elevations as exactly equal to the ratio of their ebullioscopic constants:

The problem generously provided us with the ratio of the ebullioscopic constants right at the beginning:

Therefore, by simple substitution, the ratio of their boiling point elevations must also be:

Without needing to calculate the actual molar mass of the solute, or the exact numerical value of the molality, or the individual boiling points of the solvents, we have arrived at the correct answer through pure logical deduction and an understanding of proportionalities.

Anticipating Variations

While this specific problem was straightforward because the solute was a non-electrolyte and the masses were identical, it is crucial to anticipate how examiners might twist this concept in future exams.

Variation 1: Different Solutes What if we dissolved $1\text{ g}$ of glucose in Solvent A, and $1\text{ g}$ of urea in Solvent B? Even though the masses are the same, their molar masses ($M_2$) are different. Glucose has a molar mass of $180\text{ g/mol}$, while urea has a molar mass of $60\text{ g/mol}$. This would mean the molalities would no longer be equal, and you would have to calculate the ratio of their molalities to find the final answer.

Variation 2: Electrolyte vs. Non-Electrolyte What if the solute in Solvent A was sodium chloride ($\text{NaCl}$), an electrolyte with $i=2$, while the solute in Solvent B remained a non-electrolyte with $i=1$? In this scenario, the van't Hoff factor would drastically alter the outcome. The equation for the ratio would become:

You would have to plug in $i_A=2$ and $i_B=1$, which would double the relative elevation for Solvent A compared to our original problem.

Always read the problem statement carefully. Identify the nature of the solute (electrolyte vs. non-electrolyte) to determine the van't Hoff factor, and meticulously check if the masses and molar masses allow you to assume constant molality. By mastering these underlying principles, you can confidently tackle any variation of colligative property problems!