Animated Solution for Chemistry - Basic Concepts in Chemistry: The density of NaOH solution is $1.2{\text{ g cm}}^{-3}$. The molality of this solution is ............ m. (Round off to the nearest integer) [Use : Atomic masses : Na=23.0 u, O=16.0 u, H=1.0 u, density of ${\text{H}}_2\text{O}:1.0{\text{ g cm}}^{-3}$]

The journey to mastering concentration terms often involves navigating through a maze of assumptions and approximations. This problem is a classic example of how JEE tests not just your knowledge of formulas, but your ability to make logical physical approximations when data seems "missing."

Imagine you are standing in a laboratory, holding a beaker filled with exactly $1\text{ L}$ (or $1000\text{ mL}$) of a sodium hydroxide ($\text{NaOH}$) solution. We are given the density of this solution as $1.2{\text{ g/cm}}^3$. Our ultimate goal is to find the molality of this solution.

To find the molality, we need two critical pieces of information: the number of moles of the solute ($\text{NaOH}$) and the mass of the solvent (water) strictly in kilograms.

Let's start by finding the total mass of our $1\text{ L}$ solution. Since density is the ratio of mass to volume, we can easily find the mass by multiplying the volume by the density:

So, our entire solution weighs $1200\text{ g}$. But how much of this is water, and how much is sodium hydroxide? This is where the critical assumption comes into play. The problem expects us to neglect the volume occupied by the solid sodium hydroxide. Therefore, we assume that the volume of the water is approximately equal to the total volume of the solution, which is $1000\text{ mL}$.

Using the given density of water ($1.0{\text{ g/cm}}^3$), we can find the mass of the solvent:

Now that we know the total mass of the solution is $1200\text{ g}$ and the mass of the water is $1000\text{ g}$, the difference must be the mass of our solute, sodium hydroxide:

Next, we convert this mass into moles. The molar mass of sodium hydroxide is the sum of the atomic masses of Sodium ($23$), Oxygen ($16$), and Hydrogen ($1$), which gives us $40\text{ g/mol}$.

Finally, we substitute our values into the molality formula. We have $5\text{ moles}$ of solute dissolved in $1\text{ kg}$ of solvent:

The molality of the solution is exactly $5$. This problem beautifully illustrates how a simple assumption about volume can unlock the entire calculation!