Molarity and dilutions: exercises

In Year 4 (the fourth year of Italian high school) at the liceo delle scienze umane (human sciences), the natural science programme restarts with equilibria, acids, bases, pH and redox reactions — but before all that come solutions and their concentration, between the end of September and November. The test typically arrives between the beginning and the middle of November, so between six and eight weeks from now. With only 2 hours of science a week, in class there is almost never enough time to practise this topic enough, and in fact it is the prerequisite that nobody revises later when pH and redox arrive — it is taken for granted, and those who haven’t fixed it well only find out at the test.

The problem: four ways of saying “how concentrated”

A solution always has two components: the solvent (usually water, present in the larger quantity) and the solute (the dissolved substance). “Concentration” is not a single number: it is a family of different quantities, which answer different questions and which are systematically confused in tests. Before calculating anything, it is worth telling them apart properly.

Mass percent (% m/m): how many grams of solute there are in 100 g of total solution (solute + solvent).

% m / m = \frac{m _{solute}}{m _{solution}} \times 100

Volume percent (% V/V): used for solutions between liquids, how many mL of solute there are in 100 mL of solution — it is the same idea as above, but on volumes.

Molarity (M): how many moles of solute there are in one litre of solution (not of solvent).

M = \frac{n _{solute}}{V _{solution}}

with $n$ in moles and $V$ in litres. It is the quantity used almost always in laboratory calculations and in reactions in solution.

Molality (m): how many moles of solute there are in one kilogram of solvent — not of solution.

m = \frac{n _{solute}}{m _{solvent (kg)}}

The difference between molarity and molality is the one that confuses almost the whole class: one divides by the volume of the solution, the other by the mass of the solvent alone. They are different numbers even when describing exactly the same solution, and one of the exercises below shows it with numbers.

To calculate the moles of solute from a mass you need the molar mass: if you need to revise how to read it from the periodic table and how to convert grams and moles, I have already explained it step by step in this post on the mole and stoichiometry — here I take that conversion for granted and concentrate on solutions. To read correctly the formulae of the salts that appear in the exercises (sodium chloride, sulphate, and so on) you will find the reminder in this post on nomenclature. And if you need the broader picture of the Year 3 programme for your same track, there is this post on the atom and bonds in the human sciences.

Which of the four to use, and why the teacher asks for one in particular

It is not an aesthetic choice: each quantity answers a different question, and recognising the question is half the exercise.

If the problem talks about…	You use	Why
labels, household dilutions, alcohol percentages	% m/m or % V/V	you don’t need to know the molar mass of the solute
a reaction in solution, a titration, a volume to take out	molarity	the reaction coefficients apply to moles, and the pipette measures volumes
changing temperature (boiling, freezing)	molality	volume expands with heat, the mass of the solvent does not
pollutants, traces, water hardness	ppm	percentage figures would be unreadable

The last row deserves a note, because it comes back in civic education and in earth science texts: ppm (parts per million) means milligrams of solute per kilogram of solution, that is, a million in the denominator instead of a hundred. A concentration of $5$ ppm is $0.0005%$ by mass: the same information, written in a way that is readable.

The third row is the one that explains the existence of molality, and if they ask you in the oral test it is the answer: if you heat a solution, the volume increases and therefore the molarity decreases without a single molecule of solute having left the beaker. Molality, which divides by a mass, stays identical. That is why all the formulae on boiling-point elevation and freezing-point depression use molality and not molarity.

How to prepare a solution of a given molarity

The typical problem asks: “how many grams of solute are needed to prepare $V$ litres of solution of molarity $M$ ?”. The procedure, always in the same order:

Decide the moles of solute needed, from the definition of molarity: $n = M \times V$ .
Convert it to grams with the molar mass of the solute: $m = n \times M_{molar}$ (careful: here “M” is the molar mass in g/mol, not to be confused with molarity — they are two different quantities that unfortunately share the same letter).
Weigh out that mass, dissolve it in a small amount of water and make up to volume: add water until you reach exactly $V$ in the volumetric flask, not “about”. The final volume is that of the solution, not that of the water added — the solute itself takes up space.

Dilution: $M_{1} V_{1} = M_{2} V_{2}$

When you dilute a solution, you add only solvent: the amount of solute (the moles) doesn’t change, only the total volume increases and so the concentration drops. From this comes the most used formula of this chapter:

M_{1} V_{1} = M_{2} V_{2}

where $M_{1}$ , $V_{1}$ are the molarity and volume of the starting (more concentrated) solution and $M_{2}$ , $V_{2}$ those of the final (more dilute) solution, after the addition of water. It works because the product $M \times V$ is exactly the number of moles, and that stays constant: $n_{1} = n_{2}$ .

The mistakes I see most often

Using the volume of the added solvent instead of the volume of the final solution. Molarity is always calculated on the total volume of the solution, after making up to volume — not on the volume of water you poured in.
Confusing molarity and molality. If the problem gives the mass of the solvent and asks “how many moles per kilogram of water”, it is molality; if it gives or asks for the volume of the solution, it is molarity. They are different definitions, not two names for the same thing, and in problems with density (like exercise 3 below) the two numerical values are also different from each other.
Forgetting density when you need to go from mass to volume of solution, or vice versa. If the problem gives the density of the solution, it is almost always because it is needed to link mass and volume: $V = m / d$ .
Applying $M_{1} V_{1} = M_{2} V_{2}$ to two different solutions. This formula holds only for the same solution that is diluted by adding solvent: it can’t be used for mixing two solutions of different substances, nor for a solution to which more solid solute is added.
Getting the volume units wrong. Molarity is defined in mol/litre: if the problem gives millilitres, they have to be converted into litres before dividing, not after.

Three worked exercises

1. Preparing a solution of known molarity

You want to prepare $500$ mL of a sodium hydroxide ( $NaOH$ ) solution at a concentration of $0.4$ mol/L. How many grams of $NaOH$ do you have to weigh?

Step 1 — moles needed. The volume in litres is $V = 0.500$ L:

n = M \times V = 0.4 \times 0.500 = 0.2 mol

Step 2 — grams. Molar mass $NaOH = 23 + 16 + 1 = 40$ g/mol:

m = n \times M_{molar} = 0.2 \times 40 = 8 g

Step 3 — procedure. I weigh $8$ g of $NaOH$ , dissolve them in a little distilled water inside a flask, then make up to the exact volume of $500$ mL: that is the solution at $0.4$ mol/L.

On solutions and concentration we don’t yet have a dedicated quiz in our catalogue of practice tests; you can however find the other ready-made science tests, along with those for other subjects, in the high school practice tests category — correct answer free, guided solution with your first lesson.

2. Dilution

You have $50$ mL of a hydrochloric acid solution at $2.0$ mol/L. You dilute it with water to a final volume of $250$ mL. What is the new concentration?

Converting to litres: $V_{1} = 0.050$ L, $V_{2} = 0.250$ L. I apply $M_{1} V_{1} = M_{2} V_{2}$ :

2.0 \times 0.050 = M_{2} \times 0.250

0.10 = M_{2} \times 0.250 \Rightarrow M_{2} = \frac{0.10}{0.250} = 0.4 mol/L

By diluting the volume five times (from $50$ to $250$ mL), the concentration drops by exactly five times: from $2.0$ to $0.4$ mol/L. The moles of acid, on the other hand, stay the same before and after: $0.10$ mol.

3. Mass percent, molarity and molality of the same solution

You dissolve $20$ g of sodium chloride ( $NaCl$ ) in $180$ g of water. The resulting solution has a density of $1.05$ g/mL. Calculate the mass percent, the molarity and the molality.

Mass percent. The total mass of the solution is solute plus solvent:

m_{solution} = 20 + 180 = 200 g

% m / m = \frac{20}{200} \times 100 = 10%

Molarity. First I need the volume of the solution, which I get from the density:

V_{solution} = \frac{m}{d} = \frac{200}{1.05} \approx 190.5 mL = 0.1905 L

Then the moles of $NaCl$ , with molar mass $23 + 35.5 = 58.5$ g/mol:

n_{NaCl} = \frac{20}{58.5} \approx 0.342 mol

M = \frac{0.342}{0.1905} \approx 1.79 mol/L

Molality. Here I divide the same moles by the mass of the solvent alone, in kilograms ( $180$ g $= 0.180$ kg):

m = \frac{0.342}{0.180} \approx 1.90 mol/kg

Look at the two results: $1.79$ mol/L of molarity against $1.90$ mol/kg of molality. The same solution, the same moles of solute, two different numbers — because one divides by the volume of the whole solution (solute included) and the other by the mass of the solvent alone. It is exactly the confusion that loses marks in a test when the two terms are used as synonyms.

Before the test

The overall picture of the subject, with the other topics already covered, is on sciences; the practice tests available are collected in the high school practice tests category (the questions are in Italian). The chapter that follows in the programme, acids, bases and pH, is in Acids, bases and pH in Year 4: how to calculate pH; if the starting percentages are not solid, the foundation is in Proportions and percentages in Year 1 of high school.

If you’d like a hand with this topic, the first chat is free → contact.

Solution concentration: molarity and dilutions

The problem: four ways of saying “how concentrated”

Which of the four to use, and why the teacher asks for one in particular

How to prepare a solution of a given molarity

Dilution: $M_{1} V_{1} = M_{2} V_{2}$

The mistakes I see most often

Three worked exercises

1. Preparing a solution of known molarity

2. Dilution

3. Mass percent, molarity and molality of the same solution

Before the test

Circulatory and respiratory systems: diagram and questions

Thermochemistry: enthalpy and Hess's law with exercises

Chemical kinetics: reaction rate and worked exercises

Want to put what you read into practice?

The problem: four ways of saying “how concentrated”

Which of the four to use, and why the teacher asks for one in particular

How to prepare a solution of a given molarity

Dilution: M1​V1​=M2​V2​

The mistakes I see most often

Three worked exercises

1. Preparing a solution of known molarity

2. Dilution

3. Mass percent, molarity and molality of the same solution

Before the test

Circulatory and respiratory systems: diagram and questions

Thermochemistry: enthalpy and Hess's law with exercises

Chemical kinetics: reaction rate and worked exercises

Want to put what you read into practice?

Dilution: $M_{1} V_{1} = M_{2} V_{2}$