Animated Solution for Chemistry - Basic Concepts in Chemistry: The amount of sugar (${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$) required to prepare $2\text{ L}$ of its $0.1\text{ M}$ aqueous solution is

Imagine you are standing in a pristine chemistry laboratory. In front of you is a digital weighing scale, a spatula, a bottle of pure crystalline sugar (sucrose), and a large volumetric flask. Your task is seemingly simple yet requires absolute precision: you need to prepare exactly $2\text{ L}$ of a $0.1\text{ M}$ aqueous sugar solution.

This isn't just a theoretical exercise; it is the fundamental basis of all quantitative chemistry. Whether you are formulating a life-saving intravenous drip in a hospital or standardizing reagents for a complex titration, knowing exactly how much solute to dissolve is a non-negotiable skill. Let's embark on this journey and decode the mathematics behind the molarity.

A Deep Dive

Before we touch the weighing scale, we must understand our target. What does $0.1\text{ M}$ actually mean?

Molarity ($M$) is defined as the number of moles of solute dissolved per liter of the solution. It is a measure of concentration, telling us how "crowded" the solute particles are within the solvent. The formula is elegantly simple:

Where $n$ is the number of moles of the solute, and $V$ is the volume of the solution strictly in liters. However, our laboratory scale doesn't measure in moles; it measures in grams. Therefore, we need to bridge the gap between the abstract concept of moles and the physical reality of mass. We know that the number of moles is the given mass ($w$) divided by the molar mass ($M_w$):

Substituting this into our primary equation gives us the Master Equation for this problem:

This equation is a beautiful synthesis of concentration, mass, molecular identity, and volume.

The Molar Mass of Sucrose

To use our Master Equation, we need the molar mass ($M_w$) of our solute. The chemical formula for table sugar, or sucrose, is ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$. This molecule is a behemoth compared to simple salts like $\text{NaCl}$.

To find its molar mass, we must sum the atomic masses of all its constituent atoms. We know the standard atomic masses: Carbon is $12\text{ g/mol}$, Hydrogen is $1\text{ g/mol}$, and Oxygen is $16\text{ g/mol}$. Let's do the arithmetic carefully, as a silly mistake here will derail the entire calculation:

This means that one mole of sucrose—Avogadro's number of molecules—weighs exactly $342\text{ grams}$.

Bringing It All Together

Now, we have all the pieces of the puzzle. Let's lay out our known variables: - Target Molarity ($M$) = $0.1\text{ M}$ - Target Volume ($V$) = $2\text{ L}$ - Molar Mass ($M_w$) = $342\text{ g/mol}$

We substitute these values into our Master Equation. Let's look at the raw setup:

Our goal is to isolate $w$, the mass of the sugar. We cross-multiply the denominator to the left side of the equation:

First, let's multiply the integers. $342$ times $2$ is $684$.

Finally, multiplying by $0.1$ is mathematically equivalent to dividing by $10$, which simply shifts the decimal point one place to the left.

And there we have it! To prepare this specific solution, you must carefully weigh out exactly $68.4\text{ grams}$ of sucrose, transfer it to your $2\text{ L}$ volumetric flask, and add deionized water until the bottom of the meniscus perfectly aligns with the $2\text{ L}$ calibration mark.

Traps and Practical Applications

While this problem was straightforward, competitive exams like JEE and NEET love to introduce subtle traps.

What if the volume was given as $500\text{ mL}$ instead of liters? The definition of molarity strictly requires the volume to be in liters. If you plug $500$ directly into the denominator, your answer will be off by a factor of a thousand! Always remember to convert milliliters to liters by dividing by $1000$, which modifies our Master Equation to:

Furthermore, it is crucial to remember that molarity is a temperature-dependent property. Because liquids expand when heated and contract when cooled, the volume of the solution will change with the ambient temperature of the lab. If you prepare this $0.1\text{ M}$ solution at $25^{\circ }\text{C}$ and then move it to a cold room at $10^{\circ }\text{C}$, the volume will slightly decrease, causing the molarity to slightly increase. This is why, for highly precise temperature-independent work, chemists prefer to use molality (moles of solute per kilogram of solvent) instead.

Keep these concepts clear in your mind, respect the units, and you will master the art of chemical concentrations!