Animated Solution for Chemistry - Basic Concepts in Chemistry: The molarity of ${\text{HNO}}_3$ in a sample which has density $1.4\text{ g/mL}$ and mass percentage of $63\%$ is ……… (Molecular weight of ${\text{HNO}}_3=63$)

Imagine you are standing in a chemistry lab, holding a beaker filled with a concentrated solution of nitric acid (${\text{HNO}}_3$). The label on the bottle gives you two crucial pieces of information: the density of the solution is $1.4\text{ g/mL}$, and its concentration is $63\%$ by mass ($w/w\%$). Your mission is to find its Molarity.

Molarity is the king of concentration terms, defined as the number of moles of solute dissolved per liter of the solution. To conquer this problem, we need to extract two hidden treasures from the given data: the exact moles of ${\text{HNO}}_3$ and the exact volume of the solution in liters.

When dealing with mass percentages, the most powerful weapon in your arsenal is the 100-gram assumption. Since concentration is an intensive property (it doesn't depend on how much solution you have), we can assume any amount of solution to make our math easy.

Let's assume we take exactly $100\text{ g}$ of this nitric acid solution. Why $100\text{ g}$? Because a $63\%$ mass-by-mass concentration means that out of every $100\text{ g}$ of the total solution, exactly $63\text{ g}$ is pure ${\text{HNO}}_3$. Instantly, without any complex algebra, we have isolated the mass of our solute: $W_B=63\text{ g}$.

Now that we have the mass of the solute, finding the moles is a breeze. The molecular weight ($M_w$) of ${\text{HNO}}_3$ is given as $63\text{ g/mol}$.

The formula for moles is:

Substituting our values:

Notice how beautifully the numbers cancel out! We have exactly $1\text{ mol}$ of ${\text{HNO}}_3$ in our assumed $100\text{ g}$ sample.

We have the moles, but Molarity requires the volume of the solution. This is where the density comes into play. Density bridges the gap between mass and volume. The formula is:

Rearranging this to solve for volume:

Plugging in the mass of our assumed sample ($100\text{ g}$) and the given density ($1.4\text{ g/mL}$):

Warning! This is where many students fall into a trap. The volume we just calculated is in milliliters ($\text{mL}$), but the strict definition of Molarity demands the volume to be in Liters ($\text{L}$). We must divide by $1000$ to convert it:

We now have everything we need. The master equation for Molarity ($M$) is:

Substitute the values we painstakingly derived:

The fraction flips, sending the $14$ to the numerator:

Our final answer is $14\text{ M}$.

The Direct Formula

While understanding the first-principles derivation is crucial for building a strong foundation, competitive exams like JEE demand speed. There is a direct, high-yield shortcut formula that connects Molarity, density, and mass percentage:

Let's test it with our values:

In just one line, the $63$ cancels out, $10\times 1.4$ becomes $14$, and you arrive at the exact same answer in mere seconds. Master the concept, but use the shortcut to win the race!