Clear examples of calculating molarity from mass and volume: 3 core examples (plus more)

Start with the formula (and why it actually matters)

Before we jump into the examples of calculating molarity from mass and volume: 3 examples and beyond, lock in the basic relationship:

\[ M = \frac{n}{V} \]

Where:

• M = molarity (mol/L)
• n = moles of solute (mol)
• V = volume of solution (L)

If you’re given mass instead of moles, you sneak in one more step:

\[ n = \frac{m}{M_r} \]

Where:

• m = mass of solute (g)
• M_r = molar mass (g/mol)

Combine them and you get the version you actually use in almost every example of lab calculation:

\[ M = \frac{m / M_r}{V}\quad\text{or}\quad M = \frac{m}{M_r \cdot V} \]

That’s the backbone for all the examples include in this article.

Core examples of calculating molarity from mass and volume: 3 examples everyone should master

Let’s start with three core, clean examples of calculating molarity from mass and volume: 3 examples that mirror what you’d see on exams and in basic lab prep.

Example 1: Table salt (NaCl) solution for a simple lab

Problem: You dissolve 5.85 g of sodium chloride (NaCl) in water and make the total volume up to 500.0 mL. What is the molarity of the NaCl solution?

Step 1 – Molar mass of NaCl
Na: 22.99 g/mol, Cl: 35.45 g/mol
\(M_r(\text{NaCl}) = 22.99 + 35.45 = 58.44\,\text{g/mol}\)

Step 2 – Convert mass to moles
\[ n = \frac{m}{M_r} = \frac{5.85\,\text{g}}{58.44\,\text{g/mol}} \approx 0.100\,\text{mol} \]

Step 3 – Convert volume to liters
\[ 500.0\,\text{mL} = 0.5000\,\text{L} \]

Step 4 – Calculate molarity
\[ M = \frac{n}{V} = \frac{0.100\,\text{mol}}{0.5000\,\text{L}} = 0.200\,\text{mol/L} \]

Answer: The solution is 0.200 M NaCl.

This is one of the best examples to start with because the numbers are tidy and the chemistry is familiar.

Example 2: Glucose solution for a biology experiment

Problem: A biology lab asks you to prepare 250.0 mL of a glucose (C\(_6\)H\(_{12}\)O\(_6\)) solution that ends up being 0.150 M. You accidentally weigh 6.75 g of glucose and still make the volume up to 250.0 mL. What molarity did you actually prepare?

Step 1 – Molar mass of glucose
C: 12.01 × 6 = 72.06 g/mol
H: 1.008 × 12 = 12.10 g/mol
O: 16.00 × 6 = 96.00 g/mol
Total: \(M_r \approx 180.16\,\text{g/mol}\)

Step 2 – Moles of glucose
\[ n = \frac{6.75\,\text{g}}{180.16\,\text{g/mol}} \approx 0.0375\,\text{mol} \]

Step 3 – Volume in liters
\[ 250.0\,\text{mL} = 0.2500\,\text{L} \]

Step 4 – Molarity
\[ M = \frac{0.0375\,\text{mol}}{0.2500\,\text{L}} = 0.150\,\text{M} \]

Here’s the twist: by pure luck you hit the target molarity of 0.150 M. In real labs, you rarely get that lucky, which is why these examples of calculating molarity from mass and volume matter so much.

Example 3: Sulfuric acid solution from mass of pure H₂SO₄

Problem: You have solid, pure sulfuric acid (H\(_2\)SO\(_4\)). You dissolve 9.80 g of H\(_2\)SO\(_4\) and dilute it to 400.0 mL. Find the molarity.

Step 1 – Molar mass of H₂SO₄
H: 1.008 × 2 = 2.016 g/mol
S: 32.07 g/mol
O: 16.00 × 4 = 64.00 g/mol
Total: \(M_r \approx 98.09\,\text{g/mol}\)

Step 2 – Moles of H₂SO₄
\[ n = \frac{9.80\,\text{g}}{98.09\,\text{g/mol}} \approx 0.0999\,\text{mol} \]

Step 3 – Volume in liters
\[ 400.0\,\text{mL} = 0.4000\,\text{L} \]

Step 4 – Molarity
\[ M = \frac{0.0999\,\text{mol}}{0.4000\,\text{L}} \approx 0.250\,\text{M} \]

Answer: The solution is about 0.250 M H₂SO₄.

These three core examples of calculating molarity from mass and volume: 3 examples give you the template: mass → moles → liters → molarity.

More real examples of calculating molarity from mass and volume in actual lab contexts

Once you’re comfortable with those three, it’s time for more realistic cases. These examples include buffers, medical‑style saline, and a titration scenario.

Example 4: Making a near‑physiological NaCl solution (0.90% w/v)

In medicine, 0.9% NaCl (often called “normal saline”) is standard. The CDC and hospital protocols often reference 0.9% saline for IV use, which corresponds to about 0.154 M NaCl.

Problem: You prepare a 0.90% w/v NaCl solution by dissolving 0.90 g NaCl in water and making the total volume 100.0 mL. What is the molarity?

Step 1 – Moles of NaCl
Molar mass NaCl = 58.44 g/mol (from earlier).
\[ n = \frac{0.90\,\text{g}}{58.44\,\text{g/mol}} \approx 0.0154\,\text{mol} \]

Step 2 – Volume in liters
\[ 100.0\,\text{mL} = 0.1000\,\text{L} \]

Step 3 – Molarity
\[ M = \frac{0.0154\,\text{mol}}{0.1000\,\text{L}} = 0.154\,\text{M} \]

This matches the commonly cited value for normal saline and is a nice real example of how mass‑to‑molarity math shows up in clinical and biological settings.

Example 5: Preparing a 0.10 M NaOH solution from pellets

Sodium hydroxide pellets absorb water and carbon dioxide from the air, which is a headache in analytical chemistry. But in many teaching labs, you still assume pure NaOH for practice.

Problem: You weigh 4.00 g of NaOH pellets and dissolve them in water to make 1.00 L of solution. What is the molarity?

Step 1 – Molar mass of NaOH
Na: 22.99 g/mol
O: 16.00 g/mol
H: 1.008 g/mol
Total: \(M_r \approx 40.00\,\text{g/mol}\)

Step 2 – Moles of NaOH
\[ n = \frac{4.00\,\text{g}}{40.00\,\text{g/mol}} = 0.100\,\text{mol} \]

Step 3 – Molarity
\[ M = \frac{0.100\,\text{mol}}{1.00\,\text{L}} = 0.100\,\text{M} \]

Answer: The solution is 0.100 M NaOH.

In a real analytical lab, you’d standardize this solution by titrating it against a primary standard like potassium hydrogen phthalate (KHP), which is exactly where more advanced examples of calculating molarity from mass and volume start to blend into titration calculations.

Example 6: Acetic acid in vinegar – connecting label to molarity

Household vinegar in the U.S. is often 5% acetic acid by mass. Let’s translate that into molarity, a nice bridge between chemistry class and your kitchen.

Assumption: Density of vinegar ≈ 1.01 g/mL (varies by brand; see typical food chemistry data from university sources such as USDA/education resources).

Problem: A vinegar is 5.0% w/w acetic acid (CH\(_3\)COOH). Estimate the molarity of acetic acid.

Step 1 – Mass of solution per liter
\[ 1.00\,\text{L} \times 1.01\,\text{g/mL} = 1010\,\text{g solution (approx.)} \]

Step 2 – Mass of acetic acid
5.0% w/w means 5.0 g acetic acid per 100 g solution.
\[ \frac{5.0\,\text{g}}{100\,\text{g}} \times 1010\,\text{g} \approx 50.5\,\text{g acetic acid} \]

Step 3 – Molar mass of acetic acid
C\(_2\)H\(_4\)O\(_2\): \(M_r \approx 60.05\,\text{g/mol}\)

Step 4 – Moles of acetic acid
\[ n = \frac{50.5\,\text{g}}{60.05\,\text{g/mol}} \approx 0.841\,\text{mol} \]

Step 5 – Molarity
Volume is 1.00 L, so \(M \approx 0.84\,\text{M}\).

This is one of the best examples of connecting a product label to real molarity, and it’s exactly the type of example of calculation that shows up in AP and college exams.

Example 7: Buffer component – Tris base solution

Tris(hydroxymethyl)aminomethane (Tris) is a classic buffer component in biochemistry and molecular biology. Organizations like NIH and major research institutions rely on Tris buffers in countless protocols.

Problem: You need 0.0500 M Tris base for a buffer. You accidentally prepare it by dissolving 3.03 g Tris (molar mass 121.14 g/mol) and diluting to 500.0 mL. What molarity did you actually make?

Step 1 – Moles of Tris
\[ n = \frac{3.03\,\text{g}}{121.14\,\text{g/mol}} \approx 0.0250\,\text{mol} \]

Step 2 – Volume in liters
\[ 500.0\,\text{mL} = 0.5000\,\text{L} \]

Step 3 – Molarity
\[ M = \frac{0.0250\,\text{mol}}{0.5000\,\text{L}} = 0.0500\,\text{M} \]

You actually nailed the target this time, but the point is that this type of buffer calculation is just another example of calculating molarity from mass and volume—same formula, different context.

Example 8: Diluting a solid‑based stock solution

By 2024–2025, most teaching labs are emphasizing stock solutions and dilution plans, because that’s how real labs operate. Let’s combine mass‑based prep with a later dilution.

Problem: You first prepare a 1.00 M KCl stock by dissolving 74.6 g of KCl (M\(_r\) ≈ 74.6 g/mol) to make 1.00 L of solution. Later, you pipette 25.0 mL of this stock into a flask and dilute to 250.0 mL. What is the molarity of the final solution, and how does the original mass matter?

Step 1 – Stock solution
\[ n = \frac{74.6\,\text{g}}{74.6\,\text{g/mol}} = 1.00\,\text{mol} \]
\[ M_{\text{stock}} = \frac{1.00\,\text{mol}}{1.00\,\text{L}} = 1.00\,\text{M} \]

Step 2 – Moles taken from stock
\[ V_{\text{taken}} = 25.0\,\text{mL} = 0.0250\,\text{L} \]
\[ n_{\text{taken}} = M_{\text{stock}} \times V_{\text{taken}} = 1.00\,\text{mol/L} \times 0.0250\,\text{L} = 0.0250\,\text{mol} \]

Step 3 – Final molarity after dilution
Final volume: 250.0 mL = 0.2500 L.
\[ M_{\text{final}} = \frac{0.0250\,\text{mol}}{0.2500\,\text{L}} = 0.100\,\text{M} \]

The original mass‑based preparation is still at the heart of this, but now you’ve layered dilution on top, which is how real examples in research labs usually look.

Common mistakes in examples of calculating molarity from mass and volume

Even when students understand the formula, the same errors show up over and over in examples of calculating molarity from mass and volume:

Forgetting to convert milliliters to liters

If you plug 250 instead of 0.250 into \(M = n / V\), your answer will be off by a factor of 1000. Every example of good work shows the conversion step clearly.

Using the wrong molar mass

Students often:

• Round atomic masses too aggressively.
• Forget to multiply by subscripts in the formula.

For instance, in H₂SO₄, forgetting that there are 4 oxygens instantly breaks the calculation. Checking molar masses against a periodic table or a reliable reference (many universities host them, such as Harvard’s chemistry resources) is just standard practice.

Confusing mass of solute with mass of solution

In a 5.0% solution, it matters whether that’s w/w or w/v. Many examples of calculating molarity from mass and volume in exams sneak this in. Always read the wording: grams per 100 g or grams per 100 mL?

Why these examples matter more in 2024–2025

Modern chemistry and biology labs—whether academic, clinical, or industrial—are increasingly automated, but the math hasn’t gone away. You’ll see these real examples of concentration calculations in:

• Preparing calibration standards for instruments like HPLC or ICP‑MS.
• Setting up culture media and buffers for cell biology and microbiology.
• Pharmacology and toxicology work, where concentration accuracy affects safety; organizations like Mayo Clinic and NIH routinely communicate the importance of correct dosing and solution strength.

Being fluent with examples of calculating molarity from mass and volume isn’t just about passing an exam; it’s about not sabotaging your own experiments.

Quick mental checklist for any example of mass‑to‑molarity

When you face new exam questions or lab sheets, run this in your head:

• Do I know the molar mass of the solute? If not, calculate it carefully.
• Is the mass given in grams? If not, convert.
• Is the volume in liters? If not, convert from mL to L.
• Then apply: \( M = \frac{m}{M_r \cdot V} \).

Every one of the best examples in this guide follows that script.

FAQ: examples of calculating molarity from mass and volume

Q1. Can you give another simple example of calculating molarity from mass and volume?
Yes. Suppose you dissolve 1.20 g of KBr (M\(_r\) ≈ 119.0 g/mol) in water and make the volume 250.0 mL.
Moles: 1.20 g ÷ 119.0 g/mol ≈ 0.0101 mol.
Volume: 0.2500 L.
Molarity: 0.0101 mol ÷ 0.2500 L = 0.0404 M.

Q2. How do these examples of calculating molarity relate to dilution problems?
You often start with a mass‑to‑molarity calculation to prepare a stock solution, then use the dilution relationship \(M_1V_1 = M_2V_2\). Example: after making 1.00 M KCl from mass, you dilute it to get 0.100 M, just like in Example 8.

Q3. What’s the fastest way to check if my answer makes sense?
Use estimation: if you dissolve about 0.1 mol of solute in about 1 L, your molarity should be around 0.1 M. If your final answer is 100 M or 0.000001 M, something went wrong.

Q4. Are these examples of calculations valid for solutions at any temperature?
The math is valid, but keep in mind that volume changes with temperature. For very accurate work (like analytical chemistry), solutions are prepared at a specified temperature (often 20–25 °C) because molarity depends on volume, and volume depends on temperature.

Q5. Where can I find more practice problems and data for molar mass and solutions?
Look at general chemistry resources from universities and organizations like NIH or large U.S. universities (.edu sites). Many post open‑access practice sets and periodic tables you can rely on for atomic masses.

If you can comfortably walk through these examples of calculating molarity from mass and volume: 3 examples and the extra real‑world cases, you’re in good shape for both exams and actual lab work. The formula is simple; the skill is in using it carefully, every single time.