Animated Solution for Chemistry - Basic Concepts in Chemistry: $6.02\times 10^{20}$ molecules of urea are present in $100\text{ mL}$ of its solution. The concentration of urea solution is (Avogadro constant, $N_A=6.02\times 10^{23}{\text{ mol}}^{-1}$)

Visualizing the Solution

Imagine you are standing in a chemistry lab, holding a small beaker. Inside this beaker, there is exactly $100\text{ mL}$ of a solution. But this isn't just any liquid; it's a urea solution.

If you had microscopic vision, you would be able to count exactly $6.02\times 10^{20}$ molecules of urea swimming around in that water. Our mission is to find the concentration of this solution, specifically its molarity.

Taming the Volume

Molarity is defined as the number of moles of solute per liter of solution. The formula is beautifully simple:

However, there is a catch here. The volume $V$ must strictly be in liters. We are given $100\text{ mL}$. Don't rush through this and plug in $100$ directly! We must convert it first.

Now our volume is ready for the master equation.

Counting the Uncountable

Next, we need $n$, the number of moles of urea. But the problem gave us the raw number of molecules instead. How do we bridge this gap?

This is where Avogadro's number, $N_A=6.02\times 10^{23}{\text{ mol}}^{-1}$, comes to the rescue. Just like a dozen represents $12$ items, a mole represents $6.02\times 10^{23}$ items. To find the number of moles, we simply divide our given number of molecules by Avogadro's number:

Let's substitute the values we have:

Notice how the $6.02$ cancels out perfectly? This is a classic JEE/NEET setup designed to test your conceptual clarity rather than your calculator skills.

So, we have exactly $0.001\text{ mol}$ of urea in our beaker.

Bringing It All Together

We have our moles ($n=0.001\text{ mol}$) and our volume ($V=0.1\text{ L}$). It's time to bring them back to our master equation for molarity.

Let's write this in powers of ten to avoid any silly decimal mistakes:

Converting $10^{-2}$ back to decimal form gives us $0.01\text{ M}$.

This means the concentration of our urea solution is $0.01\text{ M}$, which perfectly matches option (b). The beauty of this problem lies in its elegant cancellations and the fundamental relationship between molecules, moles, and volume.