Molarity Calculations for HSC Chemistry: c = n/V

You dissolve \(5.85\text{ g}\) of sodium chloride in water and make the final solution volume exactly \(500\text{ mL}\). Is the concentration \(5.85/500\)? It looks tempting, but those numbers measure different things: one is mass, and the other is volume.

Chemists need a way to describe how much chemical amount is packed into a certain volume of solution. That is what molarity does. Once you can move confidently between mass, moles, concentration, and volume, a large group of HSC solution calculations becomes much more manageable.

Before we calculate anything, predict this: if two solutions contain the same number of moles of solute, but one has twice the total volume, which is more concentrated?

The smaller-volume solution is more concentrated. The same amount of solute is packed into less solution.

That idea is the whole foundation of molarity.

01The mental model: amount packed into volume

Imagine \(0.50\text{ mol}\) of a solute spread through \(1.0\text{ L}\) of solution. Now imagine the same \(0.50\text{ mol}\) spread through \(2.0\text{ L}\).

The second solution has not lost any solute. It simply has more volume available, so there is less solute per litre.

This is similar to crowd density. Fifty people in a small room feels crowded. The same fifty people spread through a large hall does not. Molar concentration works in the same general way: it compares an amount with the space that amount occupies.

The analogy has a limit. Molecules are not tiny people standing still in separate patches. Particles in a solution are constantly moving and are distributed throughout the solution. The useful part of the analogy is only the idea of amount per volume.

02The molarity equation

The formal relationship is

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

where:

• \(c\) is the molar concentration, usually measured in \(\text{mol L}^{-1}\)
• \(n\) is the chemical amount of solute, measured in mol
• \(V\) is the volume of the solution, measured in L

A concentration of \(0.40\text{ mol L}^{-1}\) therefore means there are \(0.40\text{ mol}\) of solute for every litre of solution.

You may also see concentration written as \(0.40\text{ M}\), where M means molar. In HSC calculations, \(\text{mol L}^{-1}\) makes the units especially clear.

The equation can be rearranged depending on what you need:

\[ n = cV \]

and

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

Rather than memorising three unrelated equations, remember the underlying idea: concentration is amount divided by volume.

03The unit trap: litres, not millilitres

Suppose a solution contains \(0.20\text{ mol}\) in \(250\text{ mL}\). A student might calculate

\[ c = \frac{0.20}{250} \]

and obtain \(0.00080\).

The arithmetic is fine. The units are not.

If concentration is required in \(\text{mol L}^{-1}\), the volume must be in litres.

Since

\[ 1000\text{ mL} = 1\text{ L} \]

we convert millilitres to litres by dividing by 1000:

\[ 250\text{ mL} = 0.250\text{ L} \]

Now:

\[ c = \frac{0.20\text{ mol}}{0.250\text{ L}} = 0.80\text{ mol L}^{-1} \]

The factor-of-1000 error is one of the most common mistakes in concentration calculations. A useful habit is to check the volume unit before substituting anything into \(c=n/V\).

04A quick decision rule

Most molarity problems become easier if you identify what information you have and what quantity you need.

Here, \(m\) is mass in grams and \(M\) is molar mass in \(\text{g mol}^{-1}\).

That last pair matters because concentration equations work with moles, not grams.

05From mass to moles

Suppose you are given the mass of a solute rather than its chemical amount. You first use

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

where:

• \(n\) is chemical amount in mol
• \(m\) is mass in g
• \(M\) is molar mass in \(\text{g mol}^{-1}\)

Then use \(c=n/V\).

So a mass-to-concentration problem is really a two-stage path:

\[ \text{mass} \rightarrow \text{moles} \rightarrow \text{concentration} \]

Going in the other direction gives

\[ \text{concentration} \rightarrow \text{moles} \rightarrow \text{mass} \]

This chain is worth understanding rather than memorising. Concentration tells you moles per litre, while molar mass tells you grams per mole. Moles are the bridge between them.

06Concentration is based on solution volume

There is another tempting mistake hidden in the word “volume”.

Suppose you add a solute to \(500\text{ mL}\) of water. Can you automatically use \(V=0.500\text{ L}\) in \(c=n/V\)?

Not necessarily.

The equation uses the final volume of the solution, not simply the amount of solvent you started with. Dissolving a solute can change the total volume.

This is why a standard solution is prepared by dissolving the solute and then making the solution up to a precise final volume in a volumetric flask. If you want to see that process in context, read How to Prepare a Standard Solution for HSC Chemistry.

07Working backwards: concentration to mass

Sometimes the calculation starts with a desired concentration. The question then asks how much solid solute is needed.

The route reverses:

\[ n=cV \]

followed by

\[ m=nM \]

Before looking at the example, make a prediction. To make twice the volume of a solution at the same concentration, what must happen to the number of moles of solute?

It must double. Since \(n=cV\), keeping \(c\) constant while doubling \(V\) doubles \(n\).

08Finding volume from concentration

If moles and concentration are known, rearrange:

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

The answer will be in litres when \(n\) is in mol and \(c\) is in \(\text{mol L}^{-1}\).

For example, suppose \(0.0750\text{ mol}\) of a solute is needed from a \(0.500\text{ mol L}^{-1}\) solution:

\[ V=\frac{0.0750\text{ mol}}{0.500\text{ mol L}^{-1}} =0.150\text{ L} \]

which is

\[ 0.150\text{ L}\times1000=150\text{ mL} \]

So \(150\text{ mL}\) of the solution contains \(0.0750\text{ mol}\) of solute.

09What \(c=n/V\) does not tell you

The equation is powerful, but it answers one specific question: how much solute is present per volume of solution.

It does not by itself tell you:

• how strongly the solute reacts
• whether the solute fully dissociates into ions
• the concentration of each ion after dissociation
• how concentration changes during dilution or reaction
• whether a solution is saturated

For example, a \(1.0\text{ mol L}^{-1}\) glucose solution and a \(1.0\text{ mol L}^{-1}\) sodium chloride solution have the same molar concentration of solute. They do not contain the same types or numbers of dissolved particles because \(\ce{NaCl}\) dissociates into ions while glucose remains as intact molecules.

That is a separate chemical idea layered on top of molarity.

For a broader comparison of concentration units, see Ways to Measure Solution Concentration in HSC Chemistry.

10The misconception to avoid: “more moles means more concentrated”

More moles does not automatically mean a higher concentration.

Compare:

• Solution A: \(0.20\text{ mol}\) in \(0.100\text{ L}\)
• Solution B: \(0.50\text{ mol}\) in \(1.00\text{ L}\)

Solution B contains more solute overall. But:

\[ c_A=\frac{0.20}{0.100}=2.0\text{ mol L}^{-1} \]

while

\[ c_B=\frac{0.50}{1.00}=0.50\text{ mol L}^{-1} \]

Solution A is four times as concentrated.

The trap comes from looking only at the numerator. Concentration depends on the ratio of amount to volume.

11Questions and solutions

Question 1

A \(600\text{ mL}\) solution contains \(0.180\text{ mol}\) of \(\ce{KCl}\). Calculate the molar concentration of \(\ce{KCl}\).

Solution 1

The concentration is \(0.300\text{ mol L}^{-1}\).

Molar concentration is chemical amount divided by solution volume:

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

First convert the volume:

\[ 600\text{ mL}=0.600\text{ L} \]

Then substitute:

\[ c=\frac{0.180\text{ mol}}{0.600\text{ L}} =0.300\text{ mol L}^{-1} \]

So each litre of this solution would contain \(0.300\text{ mol}\) of \(\ce{KCl}\). The main trap is substituting \(600\) directly into a calculation requiring litres.

Question 2

A student needs \(200.0\text{ mL}\) of a \(0.350\text{ mol L}^{-1}\) magnesium chloride solution, \(\ce{MgCl2}\).

Calculate the chemical amount of \(\ce{MgCl2}\) required.

Solution 2

The required chemical amount is \(0.0700\text{ mol}\) of \(\ce{MgCl2}\).

Use

\[ n=cV \]

and first convert the volume:

\[ 200.0\text{ mL}=0.2000\text{ L} \]

Then:

\[ n=(0.350\text{ mol L}^{-1})(0.2000\text{ L}) =0.0700\text{ mol} \]

The litre units cancel. The result makes sense because \(0.2000\text{ L}\) is one fifth of a litre, so it should contain one fifth of the \(0.350\text{ mol}\) present per litre.

Question 3

A student dissolves \(5.30\text{ g}\) of sodium carbonate, \(\ce{Na2CO3}\), and makes the final solution volume \(500.0\text{ mL}\).

Calculate the molar concentration of the solution.

Use \(M(\ce{Na2CO3})=106.0\text{ g mol}^{-1}\).

Solution 3

The sodium carbonate concentration is \(0.100\text{ mol L}^{-1}\).

The concentration equation needs moles, so first convert the mass to chemical amount:

\[ n=\frac{m}{M} =\frac{5.30\text{ g}}{106.0\text{ g mol}^{-1}} =0.0500\text{ mol} \]

Convert the final solution volume:

\[ 500.0\text{ mL}=0.5000\text{ L} \]

Now use

\[ c=\frac{n}{V} =\frac{0.0500\text{ mol}}{0.5000\text{ L}} =0.100\text{ mol L}^{-1} \]

The key reasoning step is recognising that grams cannot be substituted directly for \(n\). The mass must first be converted to moles.

Question 4

Two students prepare sodium nitrate solutions.

Student A dissolves \(8.50\text{ g}\) of \(\ce{NaNO3}\) and makes the final volume \(250.0\text{ mL}\).

Student B dissolves \(12.75\text{ g}\) of \(\ce{NaNO3}\) and makes the final volume \(500.0\text{ mL}\).

Use \(M(\ce{NaNO3})=85.0\text{ g mol}^{-1}\).

Which solution has the greater molar concentration, and by what factor?

Solution 4

Student A’s solution has the greater molar concentration, and it is \(1.33\) times the concentration of Student B’s solution.

For Student A:

\[ n_A=\frac{8.50\text{ g}}{85.0\text{ g mol}^{-1}} =0.100\text{ mol} \]

\[ V_A=250.0\text{ mL}=0.2500\text{ L} \]

\[ c_A=\frac{0.100\text{ mol}}{0.2500\text{ L}} =0.400\text{ mol L}^{-1} \]

For Student B:

\[ n_B=\frac{12.75\text{ g}}{85.0\text{ g mol}^{-1}} =0.150\text{ mol} \]

\[ V_B=500.0\text{ mL}=0.5000\text{ L} \]

\[ c_B=\frac{0.150\text{ mol}}{0.5000\text{ L}} =0.300\text{ mol L}^{-1} \]

Compare the concentrations:

\[ \frac{c_A}{c_B} =\frac{0.400}{0.300} =1.33 \]

Student B used more \(\ce{NaNO3}\), but also used a much larger final volume. This is the trap: the solution containing more total solute is not necessarily more concentrated.

Question 5

A student wants to prepare \(300.0\text{ mL}\) of \(0.200\text{ mol L}^{-1}\) calcium chloride, \(\ce{CaCl2}\).

They calculate the required chemical amount correctly, but then argue:

“Because calcium chloride produces three ions when it dissolves, I should multiply the number of moles of \(\ce{CaCl2}\) by three before finding the required mass.”

Is the student’s reasoning correct? Calculate the required mass of anhydrous \(\ce{CaCl2}\).

Use \(M(\ce{CaCl2})=110.98\text{ g mol}^{-1}\).

Solution 5

The student’s reasoning is incorrect. The required mass is \(6.66\text{ g}\) of anhydrous \(\ce{CaCl2}\).

The stated concentration, \(0.200\text{ mol L}^{-1}\), is the concentration of \(\ce{CaCl2}\) formula units before considering dissociation. We therefore calculate the moles of \(\ce{CaCl2}\) directly:

\[ n=cV \]

Convert the volume:

\[ 300.0\text{ mL}=0.3000\text{ L} \]

Then:

\[ n=(0.200\text{ mol L}^{-1})(0.3000\text{ L}) =0.0600\text{ mol} \]

Now convert the required moles of \(\ce{CaCl2}\) to mass:

\[ m=nM =(0.0600\text{ mol})(110.98\text{ g mol}^{-1}) =6.6588\text{ g} \]

To three significant figures:

\[ m=6.66\text{ g} \]

The dissociation

\[ \ce{CaCl2 -> Ca^2+ + 2Cl^-} \]

matters if the question asks for an ion concentration. It does not mean we need three times as much calcium chloride to prepare a \(0.200\text{ mol L}^{-1}\) \(\ce{CaCl2}\) solution.

For example, assuming complete dissociation, that solution would contain \(0.200\text{ mol L}^{-1}\) of \(\mathrm{Ca}^{2+}\) and \(0.400\text{ mol L}^{-1}\) of \(\mathrm{Cl}^{-}\). That is a separate calculation based on the stoichiometric ratio in the dissociation equation.

Question 6

A laboratory bottle is labelled \(0.250\text{ mol L}^{-1}\) \(\ce{NaOH}\). A student needs exactly \(0.0500\text{ mol}\) of \(\ce{NaOH}\).

They calculate

\[ V=\frac{0.0500}{0.250}=0.200 \]

and write the answer as \(0.200\text{ mL}\).

Identify the error, give the correct volume in millilitres, and explain how the units of the equation reveal the mistake.

Solution 6

The correct volume is \(200\text{ mL}\), not \(0.200\text{ mL}\). The numerical calculation is correct, but the student has attached the wrong unit.

Using

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

gives

\[ V=\frac{0.0500\text{ mol}}{0.250\text{ mol L}^{-1}} =0.200\text{ L} \]

The unit can be checked algebraically:

\[ \frac{\text{mol}}{\text{mol L}^{-1}} =\text{L} \]

So the calculated \(0.200\) must mean \(0.200\text{ L}\).

Convert to millilitres:

\[ 0.200\text{ L}\times1000\text{ mL L}^{-1} =200\text{ mL} \]

This question exposes an important habit: units are not decorations added after a calculation. They are part of the calculation and can reveal when an answer has been interpreted incorrectly.

12Where this leads next

Once \(c=n/V\) is secure, the next step is understanding what happens when the amount of solute stays constant but the solution volume changes.

That is the logic behind dilution. If water is added, \(n\) of the original solute stays the same while \(V\) increases, so \(c\) must decrease. Following that idea leads directly to the dilution relationship \(c_1V_1=c_2V_2\), covered in Solution dilution calculations for HSC Chemistry.