What is the difference between molarity and molality?

Molarity is moles of solute per liter of solution; molality is moles of solute per kilogram of solvent. Molarity is easier to measure with glassware and is used for stoichiometry, while molality is preferred for freezing-point and boiling-point calculations because mass, unlike volume, does not change with temperature.

Do I always have to convert milliliters to liters?

Yes, if you want the answer in mol/L. Divide milliliters by 1000 before dividing into moles. Skipping this step gives an answer off by a factor of 1000, and because the result still looks like a plausible decimal, the error is easy to miss. Some chemists work in mmol/mL, which is numerically identical to mol/L, but converting to liters is the safer habit.

Does molarity change if I pour some of the solution into another beaker?

No. Molarity is an intensive property, so any portion of a well-mixed solution has the same concentration as the whole. The number of moles in that portion does change — it is proportional to the volume you took — but the ratio of moles to liters stays fixed.

Why does the procedure say to dissolve the solid before filling to the mark?

Solids take up volume as they dissolve, and that added volume is unpredictable. Dissolving in a partial volume of water first and then topping up to the etched line guarantees the final solution volume equals the flask's rated volume, which is exactly the volume assumed in the molarity calculation.

CHEM-8.2

Molarity & Solution Concentration

Learn to calculate molarity from mass or moles and solution volume, and work backward to find how much solute a target concentration requires.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Molarity & Solution Concentration, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Molarity & Solution Concentration free

What this lesson covers

A chemist almost never says "add a pinch of salt." Instead they say "prepare 250 milliliters of 0.100 M sodium chloride," and that phrase carries enough information for anyone in the world to make the exact same solution. Molarity is the language that makes solution chemistry reproducible.

In this lesson you will connect three quantities you already know how to handle — mass, moles, and volume — into a single ratio. You will calculate molarity from a mass of solute dissolved in a measured volume, and you will run the calculation in reverse to find the mass of solid you need to weigh out for a target concentration. Along the way you will see why the volume in the formula is the volume of the finished solution, not the volume of water you started with, and why that distinction shows up on nearly every lab procedure you follow this year.

What Molarity Actually Measures

Concentration answers the question "how crowded is the solute?" Molarity is the specific version chemists use most:

M = \frac{\text{moles of solute}}{\text{liters of solution}}

The unit is mol/L, abbreviated M and read aloud as "molar." A 0.50 M solution of glucose contains 0.50 mol of glucose in every liter of solution — and, just as importantly, 0.25 mol in every half liter and 0.0050 mol in every 10.0 mL. Molarity is an intensive property: it describes the solution itself, not how much of it you have. Pouring off half a beaker of 0.50 M glucose leaves you with 0.50 M glucose.

The most common conceptual mistake is reading the denominator as "liters of solvent." It is not. When solid dissolves, the particles slot into the spaces between water molecules and the total volume changes in a way that is hard to predict. That is why the definition is written in terms of the final solution volume, which you measure directly rather than calculate.

Because moles sit in the numerator, molarity also functions as a conversion factor between volume and moles. Written as a ratio you can multiply by:

\text{moles} = M \times V_{\text{(L)}}

This rearrangement is the workhorse of the rest of the unit. Any time a problem hands you a concentration and a volume of a liquid reagent, it is really handing you a number of moles in disguise.

From Grams to Molarity: The Three-Step Path

Balances measure grams; molarity needs moles. So almost every molarity calculation that starts with a solid follows the same route.

Step	What you do	Tool
1	Convert grams of solute to moles	molar mass from the periodic table
2	Convert the solution volume to liters	divide milliliters by 1000
3	Divide moles by liters	the molarity definition

Suppose you dissolve 8.50 g of potassium hydroxide, KOH, and dilute to a total volume of 400.0 mL. The molar mass is

39.10 + 16.00 + 1.01 = 56.11

g/mol, so

n = \frac{8.50\ \text{g}}{56.11\ \text{g/mol}} = 0.1515\ \text{mol}

The volume is

400.0\ \text{mL} = 0.4000

L, giving

M = \frac{0.1515\ \text{mol}}{0.4000\ \text{L}} = 0.379\ \text{M}

Two habits prevent most errors here. First, convert milliliters to liters before you divide, not after; students who divide moles by 400 get an answer 1000 times too small and rarely notice, because the number still "looks like" a concentration. Second, sanity-check the size of the result. Common laboratory solutions run roughly from 0.01 M to about 6 M. An answer of 380 M or 0.00038 M is a signal that a unit conversion slipped.

Significant figures follow the usual rules: 8.50 g has three, 400.0 mL has four, so the answer carries three.

Working Backward: Preparing a Solution of Target Concentration

Lab work usually runs the calculation in reverse. Your teacher specifies the concentration and volume, and you must decide what to weigh out. Rearrange the definition:

n = M \times V, \qquad \text{mass} = n \times \mathcal{M}

where

\mathcal{M}

is molar mass. To make 250.0 mL of 0.150 M copper(II) sulfate,

\mathrm{CuSO_4}

(molar mass 159.62 g/mol):

n = 0.150\ \text{M} \times 0.2500\ \text{L} = 0.0375\ \text{mol}

\text{mass} = 0.0375\ \text{mol} \times 159.62\ \text{g/mol} = 5.99\ \text{g}

The physical procedure matters as much as the arithmetic. You weigh 5.99 g of solid, transfer it into a 250.0 mL volumetric flask, add enough distilled water to dissolve it completely with swirling, and only then add water up to the etched calibration mark. You do not measure out 250.0 mL of water and dump the solid in — that produces slightly more than 250.0 mL of solution and therefore a concentration slightly below the target.

A second detail that trips people up: hydrated salts. If the bottle says

\mathrm{CuSO_4 \cdot 5H_2O}

, the water of hydration is part of what you weigh, so you must use the molar mass of the hydrate (249.72 g/mol), which changes the required mass to 9.36 g. Using the anhydrous molar mass for a hydrated solid gives a solution that is too dilute.

Finally, read the label of the compound carefully —

\mathrm{Na_2CO_3}

and

\mathrm{NaHCO_3}

have different molar masses and are easy to confuse under time pressure.

Ion Concentrations and Other Concentration Units

When an ionic compound dissolves, it dissociates, and the concentration of each ion depends on the formula. A 0.20 M solution of

\mathrm{CaCl_2}

contains 0.20 M

\mathrm{Ca^{2+}}

but 0.40 M

\mathrm{Cl^-}

, because each formula unit releases two chloride ions. Questions that ask for "the concentration of chloride ion" rather than "the concentration of the compound" are testing exactly this distinction, and skipping the subscript is one of the most frequent slips in this unit.

Molarity is not the only way to express concentration. Knowing what else exists keeps you from mixing formulas.

Unit	Definition	Typical use
Molarity (M)	mol solute per L solution	most lab and stoichiometry work
Molality (m)	mol solute per kg solvent	freezing- and boiling-point changes
Mass percent	$\frac{\text{mass solute}}{\text{mass solution}} \times 100$	commercial products, alloys
ppm	mg solute per L of dilute aqueous solution	trace contaminants in water

Notice that molarity uses liters of solution while molality uses kilograms of solvent. Because volume expands slightly when a liquid is warmed, molarity drifts a little with temperature; molality does not. For everyday bench chemistry near room temperature the drift is negligible, which is why molarity dominates.

Mass percent and ppm are useful when the identity of the solute is uncertain or its molar mass is irrelevant, such as reporting lead in drinking water. Molarity wins whenever you need to count particles for a reaction, because moles are what balanced equations count.

Key terms

Molarity (M).: The number of moles of solute divided by the volume of solution in liters; units of mol/L.
Solute.: The substance being dissolved, present in the smaller amount and dispersed throughout the solvent.
Solvent.: The substance doing the dissolving, present in the greater amount; water in an aqueous solution.
Solution volume.: The total volume of solute plus solvent after mixing — the quantity that goes in the denominator of molarity.
Molar mass.: The mass in grams of one mole of a substance, found by summing atomic masses from the periodic table; the bridge between grams and moles.
Volumetric flask.: Glassware calibrated to hold one precise volume at a single etched mark, used to prepare solutions of known molarity.
Stock solution.: A concentrated solution of known molarity kept on hand and later diluted to working concentrations.
Intensive property.: A property, such as concentration, whose value does not change when the amount of sample changes.

Worked example

A student dissolves 15.0 g of sodium chloride, NaCl, in distilled water and dilutes the mixture to a final volume of 250.0 mL in a volumetric flask. (a) What is the molarity of the solution? (b) How many moles of NaCl are in a 25.0 mL portion drawn from this flask? (c) What is the concentration of chloride ion?

Part (a). Start by converting grams to moles. The molar mass of NaCl is

22.99 + 35.45 = 58.44

g/mol, so

n = \frac{15.0\ \text{g}}{58.44\ \text{g/mol}} = 0.2567\ \text{mol}

Next convert the volume:

250.0\ \text{mL} \times \frac{1\ \text{L}}{1000\ \text{mL}} = 0.2500

L. Now divide.

M = \frac{0.2567\ \text{mol}}{0.2500\ \text{L}} = 1.03\ \text{M}

Three significant figures, because 15.0 g has three. The answer is a reasonable bench concentration, which is a good sign.

Part (b). Molarity is intensive, so the 25.0 mL portion is still 1.03 M. Use molarity as a conversion factor:

n = 1.03\ \frac{\text{mol}}{\text{L}} \times 0.0250\ \text{L} = 0.0257\ \text{mol}

Notice this is one tenth of the total, exactly as expected since 25.0 mL is one tenth of 250.0 mL.

Part (c). NaCl dissociates as

\mathrm{NaCl \rightarrow Na^+ + Cl^-}

, a one-to-one ratio, so the chloride concentration equals the compound concentration: 1.03 M. Had the solute been

\mathrm{MgCl_2}

, the chloride concentration would have been twice the compound's molarity.

Practice questions

What mass of potassium nitrate,

\mathrm{KNO_3}

(molar mass 101.10 g/mol), is needed to prepare 500.0 mL of a 0.250 M solution?

6.32 g
12.6 g
25.3 g
50.6 g

Answer: 12.6 g

First find moles:

n = 0.250\ \text{M} \times 0.5000\ \text{L} = 0.125

mol. Then convert to mass:

0.125\ \text{mol} \times 101.10\ \text{g/mol} = 12.6

g. The value 25.3 g comes from forgetting to halve for the 500.0 mL volume — that is the mass for a full liter. The value 6.32 g comes from dividing by two an extra time.

A solution is labeled 0.40 M

\mathrm{Al(NO_3)_3}

. What is the molar concentration of nitrate ion, and what is the total number of moles of nitrate in 750.0 mL of the solution?

Answer: 1.2 M nitrate; 0.90 mol of nitrate in 750.0 mL.

Each formula unit of

\mathrm{Al(NO_3)_3}

releases three nitrate ions, so the nitrate concentration is

3 \times 0.40 = 1.2

M. To get moles, multiply by the volume in liters:

1.2\ \text{M} \times 0.7500\ \text{L} = 0.90

mol. Students often report 0.30 mol here by using the compound's molarity instead of the ion's.

Two students each need to prepare 1.00 L of 0.100 M glucose. Student A measures 1.00 L of water in a graduated cylinder, then stirs in the weighed glucose. Student B places the weighed glucose in a 1.00 L volumetric flask, dissolves it in about 600 mL of water, then adds water to the calibration mark. Whose solution matches the target concentration, and why does the other one miss?

Answer: Student B is correct. Student A ends up with a total volume slightly greater than 1.00 L, so the concentration is slightly less than 0.100 M.

Molarity is defined per liter of solution, not per liter of solvent. Dissolved glucose occupies space, so 1.00 L of water plus the solid gives a final volume above 1.00 L. Dividing the same number of moles by a larger volume produces a concentration below the target. The volumetric flask solves this by fixing the final solution volume directly, which is why solution prep always ends with 'dilute to the mark.'

FAQ

What is the difference between molarity and molality?: Molarity is moles of solute per liter of solution; molality is moles of solute per kilogram of solvent. Molarity is easier to measure with glassware and is used for stoichiometry, while molality is preferred for freezing-point and boiling-point calculations because mass, unlike volume, does not change with temperature.
Do I always have to convert milliliters to liters?: Yes, if you want the answer in mol/L. Divide milliliters by 1000 before dividing into moles. Skipping this step gives an answer off by a factor of 1000, and because the result still looks like a plausible decimal, the error is easy to miss. Some chemists work in mmol/mL, which is numerically identical to mol/L, but converting to liters is the safer habit.
Does molarity change if I pour some of the solution into another beaker?: No. Molarity is an intensive property, so any portion of a well-mixed solution has the same concentration as the whole. The number of moles in that portion does change — it is proportional to the volume you took — but the ratio of moles to liters stays fixed.
Why does the procedure say to dissolve the solid before filling to the mark?: Solids take up volume as they dissolve, and that added volume is unpredictable. Dissolving in a partial volume of water first and then topping up to the etched line guarantees the final solution volume equals the flask's rated volume, which is exactly the volume assumed in the molarity calculation.

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