Animated Solution for Chemistry - Basic Concepts in Chemistry: What would be the molality of $20\%$ (mass/mass) aqueous solution of KI? (Molar mass of KI = $166{\text{ g mol}}^{-1}$)

Imagine you are in a chemistry lab, and you are handed a beaker labeled 20% (w/w) aqueous solution of KI. What does this label actually tell you?

The term mass by mass percentage is a very practical way of expressing concentration. It simply means that out of every 100 grams of the total solution, 20 grams is your active ingredient—the solute.

In this case, our solute is Potassium Iodide (KI). Therefore, the mass of the solute, $w_2$, is $20\text{ g}$.

But a solution is made of two parts: the solute and the solvent. Since it is an aqueous solution, the solvent is water. The mass of the solvent, $w_1$, is simply the total mass minus the solute mass. So, $w_1=100\text{ g}-20\text{ g}=80\text{ g}$.

Now that we have our raw materials quantified, we need to find the molality ($m$) of the solution.

Molality is a special concentration term because it does not depend on volume, making it completely independent of temperature changes. It is defined as the number of moles of solute dissolved per kilogram of the solvent.

The formula for molality is:

Here, $M{w}_2$ is the molar mass of the solute. The problem generously provides the molar mass of KI as $166{\text{ g mol}}^{-1}$. The factor of 1000 is there to convert our solvent mass from grams into kilograms.

Let's substitute our known values into the master equation.

We plug in $w_2=20$, $M{w}_2=166$, and $w_1=80$:

Now, let's simplify the math. We can easily see that 20 divides 80 exactly 4 times. This simplifies our expression to:

Multiplying the denominator, $166\times 4$ gives us $664$. So, we are left with:

When we perform this final division, we get approximately $1.506{\text{ mol kg}}^{-1}$.

Rounding this off to two decimal places, the molality is $1.51\text{ m}$. This perfectly matches option (b).

This problem was a straightforward application of the molality formula, but it tests a crucial skill: extracting physical quantities from a percentage concentration.

What if the examiner wanted to make this tougher? They could have given you the density of the solution and asked for the molarity. In that case, you would use the density to find the volume of the 100 g solution, and then divide the moles of solute by that volume in liters. Always be prepared to interconvert these concentration terms!