Animated Solution for Chemistry - Basic Concepts in Chemistry: A compound ${\text{H}}_2\text{X}$ with molar weight of $80\text{ g}$ is dissolved in a solvent having density of $0.4{\text{ g mL}}^{-1}$. Assuming no change in volume upon dissolution, the molality of a $3.2\text{ molar}$ solution is

Decoding the Given Information

When tackling physical chemistry problems, the first step is always to lay out your known variables. In this scenario, we are given a solution with a molarity ($M$) of $3.2{\text{ mol L}}^{-1}$. We are also provided with the density of the pure solvent, which is ${\rho }_{\text{solvent}}=0.4{\text{ g mL}}^{-1}$, and the molar mass of the solute ${\text{H}}_2\text{X}$, which is $80{\text{ g mol}}^{-1}$.

Our ultimate goal is to find the molality ($m$) of the solution. Remember the fundamental difference between these two concentration terms: molarity depends on the volume of the solution, while molality depends on the mass of the solvent. To bridge this gap, we need a clever way to relate volume and mass.

Constant Volume

The most critical phrase in this entire problem is: "Assuming no change in volume upon dissolution."

What does this physically mean? Imagine you have a beaker filled with a certain volume of pure solvent. When you add the solute ${\text{H}}_2\text{X}$ to it, the liquid level does not rise. The volume remains perfectly constant.

To make our calculations incredibly straightforward, let's assume a convenient volume for our final solution. Let $V_{\text{solution}}=1000\text{ mL}$ (or $1\text{ L}$). Because of the constant volume constraint, this immediately tells us that the initial volume of the pure solvent was also exactly $1000\text{ mL}$.

Now that we have our assumed volume, extracting the necessary components for molality is a breeze. First, let's find the moles of the solute. By definition, a $3.2\text{ M}$ solution contains $3.2\text{ moles}$ of solute in every $1\text{ L}$ of solution. Since we assumed our solution volume to be exactly $1\text{ L}$, we have exactly $3.2\text{ moles}$ of solute.

Next, we need the mass of the solvent in kilograms. We know the volume of the solvent is $1000\text{ mL}$ and its density is $0.4{\text{ g mL}}^{-1}$. Using the basic density formula ($\text{Mass}=\text{Volume}\times \text{Density}$), we can calculate the mass:

Converting this to kilograms, we get $W_{\text{solvent}}=0.4\text{ kg}$.

We now have everything we need to calculate the molality. The formula for molality is the number of moles of solute divided by the mass of the solvent in kilograms.

Substituting our values:

The final answer is $8$.

But wait, did you notice something missing from our calculation? We completely ignored the molar mass of the solute ($80{\text{ g mol}}^{-1}$)! This is a classic trap set by examiners in competitive exams like JEE. They often provide redundant information to test your conceptual confidence. If you truly understand the definitions of molarity and molality, you realize that the molar mass is entirely unnecessary here because we already had the number of moles directly from the molarity. Always trust your core concepts and only use the data that logically fits your derivation.