Animated Solution for Chemistry - Solutions: Comprehension Passage

Welcome to a classic JEE physical chemistry problem that beautifully intertwines colligative properties with ionic equilibrium and stoichiometry. At first glance, it seems like a simple plug-and-chug question about boiling point elevation. But beware! There is a hidden chemical reaction lurking in the mixture. Let's embark on this journey step by step.

Understanding Solution A

We start with Solution A, which is a $0.1$ molal aqueous solution of silver nitrate ($AgN{O}_3$). Boiling point elevation is a colligative property, which means it depends strictly on the number of solute particles dissolved in the solvent, not their identity.

Because silver nitrate is a strong electrolyte, it completely dissociates in water.

For every one mole of $AgN{O}_3$ we dissolve, we get two moles of ions. This gives us a van't Hoff factor ($i$) of $2$. Now, we can calculate the elevation in boiling point ($\Delta T_b$) using the formula:

Substituting the given values:

Since the boiling point of pure water is $100{}^{\circ }\text{C}$, the boiling point of Solution A is:

This gives us our first answer! The value of $x$ is 100.10.

Mixing the Solutions

Next, the problem states that we add an equal volume of $0.1$ molal aqueous barium chloride ($BaC{l}_2$) to Solution A to create Solution B. Here is where many students make their first mistake. They assume the molality of the ions remains the same. But think about the physical reality of mixing two solutions!

When we mix equal volumes of two dilute aqueous solutions (assuming their densities are approximately equal to that of pure water), the total mass of the solvent doubles. Molality is defined as moles of solute per kilogram of solvent. If the mass of the solvent doubles while the moles of each solute remain initially the same, the molality of each solute is exactly halved.

Therefore, in the new mixture, the initial concentrations before any reaction are:

The Precipitation Reaction

Now, let's look at the ions swimming in this new mixture. From the $0.05\text{ m}$ $AgN{O}_3$, we have $0.05\text{ m}$ of $A{g}^{+}$ and $0.05\text{ m}$ of $N{O}_3^{-}$. From the $0.05\text{ m}$ $BaC{l}_2$, we have $0.05\text{ m}$ of $B{a}^{2+}$. But wait! Each formula unit of $BaC{l}_2$ produces two chloride ions. So, the concentration of $C{l}^{-}$ is $2\times 0.05=0.10\text{ m}$.

Now, the crucial JEE trap: Silver ions ($A{g}^{+}$) and chloride ions ($C{l}^{-}$) absolutely hate being in solution together. Imagine the solution as a grand dance floor. Initially, the silver ions and nitrate ions are dancing freely. When the barium chloride solution is poured in, barium and chloride ions join the party. However, silver and chloride have an intense, unbreakable attraction. The moment they see each other, they pair up and immediately leave the dance floor, sitting out as a solid precipitate at the bottom of the beaker.

This is a stoichiometry problem now. We have $0.05\text{ m}$ of $A{g}^{+}$ and $0.10\text{ m}$ of $C{l}^{-}$. Clearly, $A{g}^{+}$ is the limiting reagent. It will be completely consumed in the reaction. The final concentration of $A{g}^{+}$ is $0$. The chloride ions will partially react. The amount remaining will be:

Finding the Difference

To find the boiling point of Solution B, we need to count all the dissolved particles that remain. Because the solid $AgCl$ precipitate is no longer dancing (dissolved), it no longer affects the mood of the party (the boiling point of the solution)!

Let's tally up the surviving ions (the spectator ions and the excess reactant):

The total effective molality ($m_{\text{total}}$) of Solution B is the sum of these concentrations:

Now, we calculate the boiling point elevation for Solution B. Since we are using the total molality of all particles, we don't need a separate van't Hoff factor (it's already accounted for by adding the individual ion concentrations).

The boiling point of Solution B is:

Finally, the question asks for the difference in the boiling points of the two solutions, expressed as $y\times 10^{-2}{}^{\circ }\text{C}$.

We can rewrite $0.025$ as $2.5\times 10^{-2}$. Therefore, the value of $|y|$ is 2.50.

This is a beautiful problem that tests your ability to stay vigilant. It's not just about knowing the formula for boiling point elevation; it's about understanding the physical chemistry of the solution at every single step.