Animated Solution for Chemistry - Basic Concepts in Chemistry: A $6.50$ molal solution of KOH (aq) has a density of $1.89{\text{ g cm}}^{-3}$. The molarity of the solution is ...... ${\text{mol dm}}^{-3}$ (Round off to the nearest integer). [Atomic masses: K : $39.0\text{ u}$, O : $16.0\text{ u}$, H : $1.0\text{ u}$]

Imagine you are in a chemistry lab, holding a beaker filled with a potassium hydroxide (KOH) solution. You know two crucial pieces of information about this solution: its molality is $6.50{\text{ mol kg}}^{-1}$ and its density is $1.89{\text{ g cm}}^{-3}$. Your mission is to find its molarity.

Before we dive into the math, let's quickly gather our tools. We need the molar mass of our solute, KOH. By adding the atomic masses of Potassium ($39.0\text{ u}$), Oxygen ($16.0\text{ u}$), and Hydrogen ($1.0\text{ u}$), we get a neat $56{\text{ g mol}}^{-1}$.

Now, how do we connect molality, density, and molarity? You could derive it from scratch by assuming $1\text{ kg}$ of solvent, finding the total mass of the solution, converting that mass to volume using density, and finally calculating molarity.

However, in the high-stakes environment of competitive exams, time is your most valuable asset. This is where our master equation comes into play:

This elegant formula directly bridges the gap between molality ($m$) and molarity ($M$), using density ($d$) and the molar mass of the solute ($M_{\text{solute}}$).

Let's carefully substitute our known values into the master equation. We plug in $6.50$ for molality, $1.89$ for density, and $56$ for the molar mass:

First, let's simplify the denominator. Multiplying $1000$ by $1.89$ gives us $1890$. Now, to eliminate the fraction, we cross-multiply:

Distributing the $6.50$ across the terms inside the parenthesis, we get:

We are almost there! Let's group all the $M$ terms on one side of the equation. Adding $364M$ to both sides yields:

Finally, dividing $12285$ by $1364$, we find our molarity:

The question asks us to round off to the nearest integer. Thus, our final molarity is $9{\text{ mol dm}}^{-3}$.

As a quick thought experiment, why do we even use both molarity and molality? Molarity depends on volume, which expands or contracts with temperature changes. Molality, on the other hand, relies strictly on mass, making it completely temperature-independent. This makes molality the undisputed champion for thermodynamics and colligative property experiments!