Q.Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL$^{-1}$?

The key idea is to calculate the molarity by determining the moles of solute and the volume of the solution.

1. Assume 100 g of the solution. Given it is 68% nitric acid by mass, the mass of $\text{HNO}_3$ (solute) is 68 g.
2. The molar mass of $\text{HNO}_3$ is $1 + 14 + (3 \times 16) = 63 \text{ g/mol}$. Moles of $\text{HNO}_3 = \frac{68 \text{ g}}{63 \text{ g/mol}} \approx 1.079 \text{ mol}$.
3. The density of the solution is $1.504 \text{ g mL}^{-1}$. Volume of solution = $\frac{\text{Mass of solution}}{\text{Density}} = \frac{100 \text{ g}}{1.504 \text{ g mL}^{-1}} \approx 66.49 \text{ mL}$. …

When we talk about the concentration of a solution, we often use different units depending on the context. This problem asks for molarity, which is a very common unit in chemistry, especially for reactions in solution. Molarity tells us how many moles of solute are present in one liter of the solution.

The information given is the mass percentage of nitric acid and the density of the solution.

• Mass percentage (68% $\text{HNO}_3$ by mass) means that for every 100 grams of the solution, 68 grams are nitric acid ($\text{HNO}_3$) and the remaining 32 grams are water. This directly gives us the mass of the solute.
• Density ($1.504 \text{ g mL}^{-1}$) relates the mass of the solution to its volume. Since molarity requires the volume of the solution, and we start with mass information (from the percentage), density is the crucial link to convert mass of solution into volume of solution.

The strategy is to pick a convenient amount of solution, calculate the moles of solute in that amount, calculate the volume of that amount of solution, and then combine these to find molarity. Assuming 100 g of solution is often the easiest starting point when a mass percentage is given, as it directly translates the percentage into grams.

Let's break down the calculation:

1. Identify the goal and definition:

   We need to find the molarity ($M$) of the $\text{HNO}_3$ solution.

   Molarity ($M$) $= \frac{\text{Moles of solute}}{\text{Volume of solution (in Liters)}}$

2. Determine the molar mass of the solute ($\text{HNO}_3$):

   The solute is nitric acid, $\text{HNO}_3$. We need its molar mass to convert the mass of $\text{HNO}_3$ into moles.

   Atomic masses: H = $1.008 \text{ g/mol}$, N = $14.007 \text{ g/mol}$, O = $15.999 \text{ g/mol}$.

   Molar mass of $\text{HNO}_3 = (1 \times 1.008) + (1 \times 14.007) + (3 \times 15.999) \text{ g/mol}$

   Molar mass of $\text{HNO}_3 = 1.008 + 14.007 + 47.997 \text{ g/mol}$

   Molar mass of $\text{HNO}_3 = 63.012 \text{ g/mol}$

3. Assume a basis for calculation and find the mass of solute:

   Since the concentration is given as 68% by mass, it's convenient to assume we have $100 \text{ g}$ of the solution.

   Mass of solution $= 100 \text{ g}$

   Mass of $\text{HNO}_3$ (solute) $= 68\%$ of $100 \text{ g} = 68 \text{ g}$

4. Calculate the moles of solute:

   Now, convert the mass of $\text{HNO}_3$ into moles using its molar mass.

   Moles of $\text{HNO}_3 = \frac{\text{Mass of } \text{HNO}_3}{\text{Molar mass of } \text{HNO}_3}$

   Moles of $\text{HNO}_3 = \frac{68 \text{ g}}{63.012 \text{ g/mol}}$

   Moles of $\text{HNO}_3 \approx 1.0791 \text{ mol}$ …

Concept: Molarity Calculation (Mass Percent → Molarity)

Step 1: Assume 100 g of solution

The acid is 68% $HNO_3$ by mass, so: mass of $HNO_3$ = 68 g, mass of water = 32 g.

Step 2: Find moles of HNO₃

Molar mass of $HNO_3$ = $1 + 14 + 48 = 63\ \text{g mol}^{-1}$

Moles = $\frac{68}{63} \approx 1.079\ \text{mol}$

Step 3: Find volume of solution

Density = $1.504\ \text{g mL}^{-1}$, mass = 100 g

Volume = $\frac{100}{1.504} \approx 66.49\ \text{mL} = 0.06649\ \text{L}$

Step 4: Calculate molarity

Molarity = $\frac{1.079}{0.06649} \approx 16.23\ \text{mol L}^{-1}$

Final Answer:

$\boxed{16.23\ \text{M}}$

Key Checkpoints for Exams …

Common Mistakes in Molality & Molarity Calculations (HNO₃ Example)

Students often confuse molality and molarity — this question is actually asking for molarity, but the same pitfalls apply to both. Here are the most frequent errors and how to avoid each.

✗ Mistake 1: Confusing % by mass with % by volume

The error:

Treating “68% nitric acid by mass” as if it means 68 g of HNO₃ per 100 mL of solution.

Why it’s wrong:

% by mass means 68 g of HNO₃ per 100 g of solution, not per 100 mL. Volume and mass are different — you need density to convert.

✓ How to avoid:

Always read the phrase carefully. If it says “by mass”, your base is grams of solution, not volume. Write it explicitly:

68 g HNO₃ in 100 g solution

✗ Mistake 2: Using the wrong mass for the solvent

The error:

When calculating molality (mol solute / kg solvent), students sometimes use the total solution mass (100 g) as the solvent mass.

Why it’s wrong:

The solvent is water, not the whole solution. If 100 g solution contains 68 g HNO₃, the solvent mass is:

✓ How to avoid:

For molality, always subtract solute mass from solution mass to get solvent mass. Then convert to kg.

✗ Mistake 3: Forgetting to convert density units properly

The error:

Using density $1.504\ \text{g mL}^{-1}$ directly without converting to g/L or remembering that 1 mL = 1 cm³.

Why it’s wrong:

Molarity uses litres of solution. If you multiply density by volume in mL, you get grams — but you need volume in L.

✓ How to avoid:

Convert density to g/L:

Or, when finding volume from mass, use:

✗ Mistake 4: Mixing up molarity and molality formulas

The error:

Using molality formula for molarity or vice versa.

Correct formulas:

• Molarity (M) = $\frac{\text{moles of solute}}{\text{volume of solution in litres}}$ …