Q.Calculate the concentration of nitric acid in moles per litre in a sample which has a density, $1.41\ g\ mL^{-1}$ and the mass per cent of nitric acid in it being 69%.

Concept: Molecular Mass Calculation and Molarity from Density

We need to convert mass percent and density into molarity (moles per litre).

Step 1: Find the mass of solution in 1 L.

Density = $1.41\ \text{g mL}^{-1}$, so mass of 1000 mL = $1.41 \times 1000 = 1410\ \text{g}$.

Step 2: Calculate the mass of $\text{HNO}_3$ in this solution.

Mass percent = 69%, so mass of $\text{HNO}_3$ = $\frac{69}{100} \times 1410 = 972.9\ \text{g}$.

Step 3: Convert mass to moles. …

Why this approach works

Molarity asks "how many moles of solute per litre of solution?" We're given two pieces of information that together let us answer this: the density tells us how much the solution weighs per unit volume, and the mass percent tells us what fraction of that weight is nitric acid. Multiply them to get grams of acid per litre, then convert grams to moles using the molar mass.

The key insight is that density bridges the gap between mass-based composition (mass percent) and volume-based concentration (molarity).

Step-by-step calculation

1. Find the mass of 1 litre of solution

The density is $1.41\ \text{g mL}^{-1}$. Since $1\ \text{L} = 1000\ \text{mL}$:

2. Calculate the mass of nitric acid in that litre

The solution is 69% nitric acid by mass, meaning 69 g of $\text{HNO}_3$ in every 100 g of solution:

3. Determine the molar mass of nitric acid

4. Convert mass to moles

5. Express as molarity

Since we calculated moles in exactly 1 litre of solution:

$\text{Molarity} = \frac{15.46\ \text{mol}}{1\ \text{L}} = 15.46\ \text{mol L}^{-1}$ …

Method: Density–Mass Percent–Molarity Conversion

This is a unit conversion problem that uses the definition of molarity and the given density and mass percent.

Steps

1. Assume a convenient volume

   Take 1 litre ($1000\ mL$) of the solution. This makes the final molarity calculation direct.

2. Find the mass of the solution

   Using density:

3. Find the mass of pure HNO₃ Mass percent = 69%, so:

4. Convert mass of HNO₃ to moles Molar mass of HNO₃ = $1 + 14 + 48 = 63\ g\ mol^{-1}$

5. Calculate molarity Molarity = moles per litre of solution:

Why this works …

Common Mistakes in Molecular Mass & Concentration Calculation

Students often lose marks on this exact problem. Here are the most frequent errors — and how to avoid each one.

Mistake 1: Confusing mass per cent with actual mass

The error:

Taking 69% as 69 g of HNO₃ in 100 g of solution — that part is correct — but then forgetting that the total solution mass is 100 g, not 1 L or 1 mL.

How to avoid:

Always write down:

69% by mass means 69 g HNO₃ in 100 g solution.

Do not skip writing the "100 g solution" part — it anchors your next step.

Mistake 2: Using density incorrectly (units mismatch)

The error:

Density is $1.41\ g\ mL^{-1}$. Students multiply $1.41 \times 69$ or use density as if it's $g\ L^{-1}$.

How to avoid:

Convert density to $g\ L^{-1}$ first:

Now you know: 1 L of solution weighs 1410 g.

Mistake 3: Forgetting to find mass of HNO₃ in 1 L

The error:

Jumping straight to moles without first finding how much HNO₃ is actually present in 1 L.

How to avoid:

Use the mass per cent as a conversion factor:

Calculate:

Mistake 4: Using wrong molar mass of HNO₃

The error:

Using 63 g/mol (correct) but forgetting to add oxygen atoms properly, or using 64 g/mol.

How to avoid:

Calculate molar mass step by step:

Always write the sum explicitly — never do it mentally.

Mistake 5: Dividing mass by molar mass incorrectly

The error:

Writing $\frac{972.9}{63}$ but making a decimal slip (e.g., getting 15.4 instead of 15.44).

How to avoid:

Do the division carefully:

Check reasonableness: 69% HNO₃ is concentrated — answer should be around 15–16 M. If you get 1.5 M, you misplaced a decimal.

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