Best examples of calculating molality of a solution (step‑by‑step)

Starting with simple examples of calculating molality of a solution

Let’s skip the theory-first approach and jump straight into realistic numbers. Molality (symbol m) is defined as:

\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]

Notice: kilograms of solvent, not total solution mass and not volume. That twist is exactly why walking through detailed examples of calculating molality of a solution is so helpful.

Example of a basic salt solution (kitchen-style)

Imagine you dissolve 10.0 grams of table salt (NaCl) in 250.0 grams of water.

• Mass of solute (NaCl): 10.0 g
• Mass of solvent (water): 250.0 g = 0.250 kg
• Molar mass of NaCl: about 58.44 g/mol

Moles of NaCl:

[
\text{moles NaCl} = \frac{10.0\,\text{g}}{58.44\,\text{g/mol}} \approx 0.171\,\text{mol}
]

Molality:

[
m = \frac{0.171\,\text{mol}}{0.250\,\text{kg}} \approx 0.684\,\text{mol/kg}
]

So this everyday kitchen mixture is about 0.68 m NaCl. This is one of the cleanest examples of calculating molality of a solution: everything is given directly in grams, and the solvent is clearly water.

Real examples: sports drink sugar concentration

Now a more realistic, slightly messy case. Suppose you’re reverse‑engineering a sports drink recipe. You dissolve 45.0 grams of glucose (C₆H₁₂O₆) in 355 grams of water (roughly a 12‑oz drink).

• Mass of solute (glucose): 45.0 g
• Mass of solvent (water): 355 g = 0.355 kg
• Molar mass of glucose: about 180.16 g/mol

Moles of glucose:

[
\text{moles glucose} = \frac{45.0\,\text{g}}{180.16\,\text{g/mol}} \approx 0.250\,\text{mol}
]

Molality:

[
m = \frac{0.250\,\text{mol}}{0.355\,\text{kg}} \approx 0.704\,\text{mol/kg}
]

So the drink is about 0.70 m in glucose. This is a nice example of calculating molality of a solution when you’re thinking about energy drinks, sports hydration, or nutrition labels.

If you’re curious how this compares to real products, nutrition and carbohydrate guidelines from sources like the NIH and CDC often discuss sugar intake in grams; molality lets you connect those grams back to physical properties like freezing point and osmotic pressure.

Antifreeze in a car radiator: one of the best examples

Automotive antifreeze is a classic textbook case because temperature changes matter. Assume you mix 1.50 kg of ethylene glycol (C₂H₆O₂) with 2.50 kg of water in a radiator.

• Mass of solute (ethylene glycol): 1.50 kg = 1500 g
• Mass of solvent (water): 2.50 kg
• Molar mass of ethylene glycol: about 62.07 g/mol

First convert solute mass to moles:

[
\text{moles EG} = \frac{1500\,\text{g}}{62.07\,\text{g/mol}} \approx 24.17\,\text{mol}
]

Molality:

[
m = \frac{24.17\,\text{mol}}{2.50\,\text{kg}} \approx 9.67\,\text{mol/kg}
]

So this mixture is about 9.7 m ethylene glycol. That high molality explains why the freezing point of the coolant drops so much, helping your car survive sub‑freezing winters.

This is one of the best examples of calculating molality of a solution to connect the math with a real‑world outcome: freezing point depression, which you’ll often see discussed in physical chemistry and engineering courses.

Lab‑style example: NaCl and freezing point depression

In many general chemistry labs, you measure freezing point depression to estimate molar mass. Let’s say you dissolve 3.00 grams of NaCl in 100.0 grams of water and want the molality for your calculations.

• Mass of solute (NaCl): 3.00 g
• Mass of solvent (water): 100.0 g = 0.1000 kg
• Molar mass of NaCl: 58.44 g/mol

Moles of NaCl:

[
\text{moles NaCl} = \frac{3.00\,\text{g}}{58.44\,\text{g/mol}} \approx 0.0513\,\text{mol}
]

Molality:

[
m = \frac{0.0513\,\text{mol}}{0.1000\,\text{kg}} = 0.513\,\text{mol/kg}
]

So the solution is 0.513 m NaCl. This kind of experiment is covered in many college chemistry courses; for a deeper dive into colligative properties, you can check open materials from universities like MIT OpenCourseWare or Harvard’s chemistry resources.

Examples include very dilute solutions: tap water contaminants

Real water systems often contain trace contaminants measured in parts per million or even lower. Molality is handy because it’s independent of temperature and volume.

Suppose a groundwater sample contains 5.0 mg of lead(II) nitrate [Pb(NO₃)₂] in 1.00 kg of water. You want the molality of Pb(NO₃)₂.

• Mass of solute: 5.0 mg = 0.0050 g
• Mass of solvent: 1.00 kg
• Molar mass of Pb(NO₃)₂: about 331.2 g/mol

Moles of Pb(NO₃)₂:

[
\text{moles Pb(NO}_3\text{)}_2 = \frac{0.0050\,\text{g}}{331.2\,\text{g/mol}} \approx 1.51 \times 10^{-5}\,\text{mol}
]

Molality:

[
m = \frac{1.51 \times 10^{-5}\,\text{mol}}{1.00\,\text{kg}} = 1.51 \times 10^{-5}\,\text{mol/kg}
]

So the molality is 1.5 × 10⁻⁵ m. This is a good example of calculating molality of a solution where the numbers are tiny but still meaningful, similar to the way agencies like the EPA track contaminants in drinking water.

Pharmaceutical example: saline solution for medical use

Normal saline used in medicine is about 0.9% NaCl by mass. Suppose a hospital pharmacy prepares a batch by dissolving 9.0 grams of NaCl in 991 grams of water to make roughly 1.0 kg of solution. We’ll treat the 991 g as the solvent mass.

• Mass of solute (NaCl): 9.0 g
• Mass of solvent (water): 991 g = 0.991 kg
• Molar mass of NaCl: 58.44 g/mol

Moles of NaCl:

[
\text{moles NaCl} = \frac{9.0\,\text{g}}{58.44\,\text{g/mol}} \approx 0.154\,\text{mol}
]

Molality:

[
m = \frac{0.154\,\text{mol}}{0.991\,\text{kg}} \approx 0.155\,\text{mol/kg}
]

So normal saline is about 0.155 m NaCl. This is lower than the molarity (roughly 0.154 M) but very close because the density is near 1 g/mL. Resources like Mayo Clinic and MedlinePlus often mention saline in clinical contexts; molality gives you a more thermodynamically tidy way to think about its concentration.

Temperature comparison: why molality beats molarity in some cases

One of the best examples of examples of calculating molality of a solution comes from comparing hot and cold solutions. Imagine you dissolve the same amount of urea (CH₄N₂O) in 500 g of water, then heat the solution.

You add 30.0 g of urea to 500.0 g of water.

• Mass of solute: 30.0 g
• Mass of solvent: 500.0 g = 0.500 kg
• Molar mass of urea: about 60.06 g/mol

Moles of urea:

[
\text{moles urea} = \frac{30.0\,\text{g}}{60.06\,\text{g/mol}} \approx 0.500\,\text{mol}
]

Molality:

[
m = \frac{0.500\,\text{mol}}{0.500\,\text{kg}} = 1.00\,\text{mol/kg}
]

Now here’s the key point: this 1.00 m value does not change when you heat the solution, because the mass of solvent stays the same. But the molarity (moles per liter of solution) will change as the solution expands with temperature.

This makes molality especially useful in thermodynamics and physical chemistry, where you need a concentration unit that doesn’t wiggle around every time the temperature shifts. Many modern physical chemistry texts (check your local university’s .edu library) emphasize this difference with similar examples of calculating molality of a solution at different temperatures.

Gas absorption example: CO₂ in water under pressure

Consider a carbonated beverage where 2.20 grams of CO₂ are dissolved in 1.00 kg of water under pressure.

• Mass of solute (CO₂): 2.20 g
• Mass of solvent (water): 1.00 kg
• Molar mass of CO₂: about 44.01 g/mol

Moles of CO₂:

[
\text{moles CO}_2 = \frac{2.20\,\text{g}}{44.01\,\text{g/mol}} \approx 0.0500\,\text{mol}
]

Molality:

[
m = \frac{0.0500\,\text{mol}}{1.00\,\text{kg}} = 0.0500\,\text{mol/kg}
]

So the solution is 0.0500 m CO₂. This kind of setup shows up in environmental chemistry and chemical engineering when modeling gas solubility. Again, it’s another concrete example of calculating molality of a solution where the solvent mass is easy to track, but the volume might be less obvious.

Common mistakes when working through examples of calculating molality of a solution

When you practice more examples of examples of calculating molality of a solution, you’ll see the same error patterns:

• Using total solution mass instead of solvent mass. Always isolate the solvent. If you’re given “100 g solution, 10 g solute,” remember the solvent is 90 g.
• Mixing up molarity and molality. Molarity uses liters of solution; molality uses kilograms of solvent. If you see a volume in milliliters, that’s a hint you might be dealing with molarity instead.
• Forgetting to convert grams to kilograms for the solvent. This is probably the single most common point loss on exams.
• Rounding molar masses too aggressively. For most coursework, two decimal places on molar mass are fine, but don’t round 58.44 g/mol down to 58 g/mol unless you know your instructor doesn’t care.

When you’re stuck, check whether your setup matches the pattern in the earlier real examples of calculating molality of a solution: moles on top, kilograms of solvent on the bottom, and all units clearly labeled.

Quick mental checks using real examples

You can often sanity‑check your answer just by comparing it to these earlier cases:

• If you add a few grams of solute to hundreds of grams of water, you should expect a molality well under 1 m (like the sports drink and 0.9% saline examples).
• If you add hundreds of grams or more of solute to a similar mass of solvent, you’ll get multi‑molal solutions (like the antifreeze example at about 9.7 m).
• Extremely small masses of solute in kilograms of solvent will give you tiny molalities, often in scientific notation (like the lead nitrate groundwater example).

If your answer is wildly outside these ballparks, treat that as a red flag and re‑run the calculation.

FAQ: examples of common student questions about molality

Q1. Can you give another quick example of converting grams to molality?
Sure. Suppose you have 5.00 g of KCl dissolved in 200.0 g of water. Molar mass of KCl ≈ 74.55 g/mol. Moles KCl = 5.00 / 74.55 ≈ 0.0671 mol. Solvent mass = 0.200 kg. Molality = 0.0671 / 0.200 ≈ 0.336 m. This is a compact example of calculating molality of a solution that you can do almost in your head once you’re comfortable.

Q2. Why do textbooks prefer molality in colligative property examples?
Because colligative properties (freezing point depression, boiling point elevation, osmotic pressure) depend on the number of particles, and molality doesn’t change with temperature. As you saw in the urea and antifreeze examples of calculating molality of a solution, molality stays fixed even if the solution expands or contracts.

Q3. Where can I find more worked examples of molality problems?
Many general chemistry courses post practice sets online. Look for .edu resources or open course materials. Sites like Khan Academy and university chemistry departments often include step‑by‑step examples of molality, molarity, and related concentration units.

Q4. Is molality used in real labs, or is it just a classroom topic?
It’s very much used in real work, especially in physical chemistry, thermodynamics, and solution property measurements. When researchers publish data on activity coefficients or colligative properties, molality is often the preferred unit. That’s why so many real examples of calculating molality of a solution show up in research papers and technical reports.

If you can comfortably re‑work each of the eight real examples here—kitchen salt, sports drink, antifreeze, lab NaCl, groundwater contaminant, saline, urea, and carbonated water—you’ve already covered the main patterns that show up in homework, quizzes, and exams. From there, any new problem is just a small variation on a familiar theme.