Animated Solution for Chemistry - Basic Concepts in Chemistry: $6.023\times 10^{22}$ molecules are present in $10\text{ g}$ of a substance 'x'. The molarity of a solution containing $5\text{ g}$ of substance 'x' in $2\text{ L}$ solution is ...... $\times 10^{-3}$.

A Journey through Moles and Molarity

Imagine you are a chemist in a laboratory, handed a mysterious vial containing a substance labeled simply as 'x'. Your task is to prepare a solution of a specific concentration, but there's a catch—you don't know what the substance is, nor do you know its molar mass. However, you are given a crucial piece of detective evidence: $10\text{ g}$ of this substance contains exactly $6.023\times 10^{22}$ molecules. Let's break down how we can use this clue to solve the entire puzzle.

Finding the Molar Mass

Before we can even think about making a solution, we need to understand the fundamental nature of substance 'x'—specifically, its molar mass. The mole concept is our ultimate translation tool here. It allows us to bridge the microscopic world of molecules with the macroscopic world of grams.

The master equation that connects these realms is:

Here, $N$ is the number of molecules, $N_A$ is Avogadro's number ($6.023\times 10^{23}$), $w$ is the given mass, and $M$ is the molar mass we desperately need. Let's substitute our known values into this equation:

Notice how beautifully the numbers are designed to work together! The $6.023$ cancels out perfectly, leaving us with a simple ratio of powers of ten:

Solving for $M$, we find that the molar mass of substance 'x' is $100\text{ g/mol}$. The first major hurdle is cleared!

Calculating Molarity

Now that we know the molar mass, we can proceed to the actual experiment. We take $5\text{ g}$ of substance 'x' and dissolve it in enough solvent to make exactly $2\text{ L}$ of solution. Our goal is to find the molarity of this new mixture.

Molarity ($M$) is defined as the number of moles of solute per liter of solution. We can expand this definition using the mass and molar mass:

Let's plug in our new set of parameters: $w=5\text{ g}$, $M=100\text{ g/mol}$, and $V=2\text{ L}$.

Formatting the Answer

We have successfully calculated the molarity as $0.025\text{ M}$. However, JEE questions often test your ability to follow specific formatting instructions. The question asks for the answer in the form of an integer multiplied by $10^{-3}$.

To achieve this, we simply shift the decimal point three places to the right:

Therefore, the integer value that fills the blank is $25$. By systematically applying the mole concept and the definition of molarity, we've turned a cryptic set of numbers into a clear, confident answer.