Animated Solution for Chemistry - Basic Concepts in Chemistry: The mole fraction of a solvent in aqueous solution of a solute is $0.8$. The molality (in ${\text{mol kg}}^{-1}$) of the aqueous solution is

Welcome, future engineers and doctors! Today, we are going to dissect a beautiful problem from JEE Main 2019. At first glance, this question looks like a simple plug-and-chug exercise, but it hides a very elegant conceptual bridge between two fundamental concentration terms: mole fraction and molality.

Let's look at the first clue: "aqueous solution". This single word is doing a lot of heavy lifting. In chemistry, whenever you see the word "aqueous," it is a direct, non-negotiable statement that the solvent is water ($H_{2}O$). Why does this matter? Because it secretly hands us the molar mass of the solvent, $M_{\text{solvent}}=18{\text{ g mol}}^{-1}$. Without this hidden piece of data, the problem would be mathematically impossible to solve.

The problem states that the mole fraction of the solvent is $0.8$. Let's denote the solvent as component $A$ and the solute as component $B$. So, ${\chi }_A=0.8$.

What does this really mean? Mole fraction is simply a game of ratios. If you had exactly $1$ mole of this solution, $0.8$ moles would be water, and the remaining $0.2$ moles would be the solute. This is because, in any binary solution, the sum of the mole fractions of all components must equal exactly $1$:

Therefore, the mole fraction of the solute is:

Bridging Molality and Mole Fraction

We need to find the molality ($m$). I know you can just memorize the formula, but deriving it gives you immense analytical power. Let's build the bridge from scratch.

By definition, the mole fractions are:

If we take the ratio of the mole fraction of the solute to the mole fraction of the solvent, the denominators cancel out beautifully:

Now, we know that the number of moles of the solvent ($n_A$) is equal to its given mass ($W_A$) divided by its molar mass ($M_A$). So, $n_A=\frac{W_A}{M_A}$. Substituting this into our ratio gives:

Let's rearrange this to isolate the term $\frac{n_B}{W_A}$:

Now, recall the definition of molality ($m$). Molality is the number of moles of solute ($n_B$) dissolved per kilogram of solvent. If $W_A$ is in grams, we must multiply by $1000$ to convert it to kilograms:

Substitute our rearranged ratio into this definition, and boom! We have our master equation:

This equation is a high-yield tool for JEE and NEET. It directly connects a temperature-independent ratio (mole fraction) to another temperature-independent concentration term (molality).

Now, let's plug in the numbers. We have ${\chi }_B=0.2$, ${\chi }_A=0.8$, and $M_A=18{\text{ g mol}}^{-1}$.

The arithmetic here is straightforward, but let's be careful to avoid silly mistakes. The ratio $\frac{0.2}{0.8}$ simplifies perfectly to $\frac{1}{4}$.

Dividing $1000$ by $72$ gives us:

And there we have it! The answer matches option (c) perfectly. This problem teaches us to read carefully, extract hidden data like the molar mass of water from the word "aqueous," and seamlessly transition between different concentration terms. Keep practicing these derivations, and soon, you'll be able to see through the matrix of physical chemistry problems!