Concentration of Solutions (Molarity) Calculations

Key Concepts

Concentration of a solution refers to the amount of solute dissolved in a given solvent.

term	description	example
solute	substance that dissolves	solid sodium chloride, NaCl(s)
solvent	substance that enables solute to dissolve	liquid water, H₂O(l)
solution	mixture of solute dissolved in solvent	sodium chloride dissolved in water, NaCl(aq)

Concentration of a solution can be given in moles of solute per litre of solution (mol L^-1 or mol/L or M).⁽¹⁾

solute solvent solution

units of measurement moles
mol litres
L moles per litre
mol L^-1 (mol/L or M)

Molarity is the term used to describe a concentration given in moles per litre.

Molarity has the units mol L^-1 (or mol/L or M).

Molarity, concentration in mol/L or mol L^-1, is given the symbol c (sometimes M).
For a 0.01 mol L^-1 HCl solution we can write :
[HCl] = 0.01 mol L^-1 (concentration implied by square brackets around formula)
or
c(HCl) = 0.01 mol L^-1 (c stands for concentration, formula given in round brackets or parentheses)

	solute	solvent	solution
units of measurement	moles mol	litres L	moles per litre mol L^-1 (mol/L or M)

Equation (formula or expression) to calculate the molarity of a solution (concentration in mol L^-1) is

c = n ÷ V
where	c = concentration of solution in mol L^-1 (mol/L or M), n = moles of substance being dissolved (moles of solute), V = volume of solution in litres (L)

This equation (formula or expression) can be re-arranged to find:
moles of solute given molarity and volume of solution:
n = c × V
volume of solution given moles of solute and molarity:
V = n ÷ c

Please do not block ads on this website.
No ads = no money for us = no free stuff for you!

Molarity Concepts

Molarity is the term given for concentrations that are given in units of moles per litre (mol L^-1 or mol/L or M)

Consider the diagram on the right.

The black box represents a 1 litre container.

The white space in the box represents the volume occupied by the solvent.

Each blue X represents 1 mole of sugar molecules.

How many moles of sugar molecules are shown in this container? 1 mole
What is the volume of the container in litres? 1 L

Molarity, concentration in mol L^-1, tells us how many moles of solute are in 1 L of solution.

What is the concentration of sugar molecules in the solution?
There is 1 mole of sugar molecules per 1 litre of solution.
That is 1 mole per litre, or 1 mol L^-1
Concentration of sugar molecules = 1 mol L^-1 (1 mol/L or 1 M)

concentration of sugar solution = moles of sugar ÷ volume of solution (in L)

Consider the diagram on the right.

The red box represents a 0.5 litre container.

The white space in the box represents the volume occupied by the solvent.

Each blue X represents 1 mole of sugar molecules.

How many moles of sugar molecules are shown in this container? 1 mole
What is the volume of the container in litres? 0.5 L

What is the concentration of sugar molecules in the solution?
There is 1 mole of sugar molecules per 0.5 litre of solution.

Molarity, concentration in mol L^-1, tells us how many many moles of solute are in 1 L of solution.
If we divide both the moles of sugar molecules and the volume by the volume, that is divide by 0.5:
There are (1 ÷ 0.5 = 2) moles of sugar molecules per (0.5 ÷ 0.5 = 1) litres of solution
There are 2 moles of sugar molecules per 1 litre of solution
That is 2 moles per litre, or 2 mol L^-1
Concentration of sugar molecules = 2 mol L^-1

concentration of sugar solution = moles of sugar ÷ volume of solution ( in L)

We can check this equation for our example:
moles sugar = 1 mol
volume = 0.5 L
concentration = moles ÷ volume = 1 ÷ 0.5 = 2 mol L^-1 ( 2 mol/L or 2 M)

Do you know this?

Join AUS-e-TUTE!

Play the game now!

Molarity Equation

The molarity of a solution is given by the equation :

n
V

c = concentration in mol L^-1
n = moles of solute in mol
V = volume of solution in L

If you know the moles of solute in a solution, and, you know the volume of the solution, you can calculate the concentration of the solution in mol L^-1 using c = n ÷ V

But what if you know the concentration of the solution in mol L^-1 and the volume of the solution in L, can you calculate how many moles of solute are present in the solution?
Yes! You just need to rearrange the equation by multiplying both sides of the equation by V (the volume of the solution):

c × V	=	n × V V
c × V	=	n

You can even calculate the volume of the solution if you know how many moles of solute are present and its concentration in mol L^-1.
Just rearrange the equation above by dividing both sides by c (concentration of solution in mol L^-1)

c × V c	=	n c
V	=	n c

Do you understand this?

Join AUS-e-TUTE!

Take the test now!

Examples of Molarity Calculations with Worked Solutions

1. Calculating Concentration (c = n ÷ V)

Calculate the concentration in mol L^-1 (molarity) of a sodium chloride solution containing 0.125 moles sodium chloride in 0.50 litres of solution.

What is the question asking you to calculate?
c = molarity (concentration in mol L^-1) of solution = ? mol L^-1

What information has been given in the question?
Extract the data from the question
n = moles of solute = 0.125 mol
V = volume of solution = 0.50 L

What is the relationship between what you know and what you need to find?
Write the equation:
c = n ÷ V

Substitute the values into the equation for molarity and solve:
[NaCl_(aq)] = c(NaCl_(aq)) = 0.125 ÷ 0.50 = 0.25 mol L^-1 (or 0.25 mol/L or 0.25 M)

2. Calculating Moles of Solute (n = c × V)

Calculate the moles of copper sulfate in 250.00 mL of 0.020 mol L^-1 copper sulfate solution.

What is the question asking you to calculate?
n = moles of solute = ? mol

What information has been given in the question?
Extract the data from the question:
c = concentration (molarity) of solution = 0.020 mol L^-1
V = volume of solution = 250.00 mL = 250.00 ÷ 1000 = 250.00 x 10^-3 L = 0.25 L (since there are 1000 mL in 1 L)

What is the relationship between what you know and what you need to find?
Write the equation:
n = c × V

Substitute the values into the molarity equation and solve:
n = 0.020 × 0.25 = 0.0050 mol

3. Calculating Volume of Solution (V = n ÷ c)

Calculate the volume of a 0.80 mol L^-1 potassium bromide solution containing 1.60 moles of potassium bromide.

What is the question asking you to calculate?
V = volume of solution in litres = ? L

What information has been given in the question?
Extract the data from the question:
n = moles of solute = 1.60 mol
c = molarity (concentration in mol L^-1) of solution = 0.80 mol L^-1

What is the relationship between what you know and what you need to find?
Write the equation:
V = n ÷ c

Substitute the values into the equation and solve:
V = 1.60 ÷ 0.80 = 2.00 L

Can you apply this?

Join AUS-e-TUTE!

Do the drill now!

Problem Solving for Molarity Calculations

The Problem:

Chris the Chemist has been given a 250.00 mL volumetric flask and asked to use it to make an aqueous solution of sodium chloride, NaCl, with a concentration of 0.10 mol L^-1 for a corrosion experiment. The sodium chloride, NaCl, is available as white crystals. Determine the mass in grams of sodium chloride that Chris the Chemist will need to weigh out.

Solving the Problem

Using the StoPGoPS model for problem solving:

STOP!	State the question.	What is the question asking you to do? Determine (calculate) the mass of sodium chloride in grams. m(NaCl) = mass of sodium chloride = ? g
PAUSE!	Plan.	What information (data) have you been given? solute is NaCl(s) solvent is water (we know this because we are told Chris makes an aqueous solution) V = volume of NaCl(aq) = 250.00 mL c = concentation (molarity) of NaCl(aq) = 0.10 mol L^-1 What is the relationship between what you know and what you need to find out? Calculate moles of NaCl in solution solution n = c × V c = 0.10 mol L^-1 V = 250.00 L Convert volume in mL to volume in L: V(L) = V(mL) ÷ 1000 Assume the temperature of the laboratory is the same as the temperature required by the volumetric flask (eg 25°C). Assume the water used to make up the solution does not contain any NaCl. n(NaCl) = c × V Calculate mass of NaCl(s) to be used n = mass ÷ molar mass Use the Periodic Table to find the relative atomic masses for (i) M_r(Na) (ii) M_r(Cl) Find the molar mass, M, of NaCl: M(NaCl) = M_r(Na) + M_r(Cl) Assume the NaCl(s) that is to be used is 100% pure (no impurities) m(NaCl) = n(NaCl) × M(NaCl)
GO!	Go with the Plan.	Step 1: Calculate moles of NaCl in solution solution c = 0.10 mol L^-1 V = 250.00 L Convert volume in mL to volume in L: V(L) = V(mL) ÷ 1000 = 250.00 ÷ 1000 = 0.25 L Assume the temperature of the laboratory is the same as the temperature required by the volumetric flask (eg 25^oC). Assume the water used to make up the solution does not contain any NaCl. n(NaCl) = c × V = 0.10 × 0.25 = 0.025 mol Step 2: Calculate mass of NaCl(s) to be used Use the Periodic Table to find the relative atomic masses for M_r(Na) = 22.99 M_r(Cl) = 35.45 Find the molar mass, M, of NaCl: M(NaCl) = M_r(Na) + M_r(Cl) = 22.99 + 35.45 = 58.44 g mol^-1 Assume the NaCl(s) that is to be used is 100% pure (no impurities) m(NaCl) = n(NaCl) × M(NaCl) = 0.025 × 58.44 = 1.46 g
PAUSE!	Ponder Plausability.	Have you answered the question that was asked? Yes, we have calculated the mass of NaCl that needs to be used to make the solution. Is your solution to the question reasonable? Let's work backwards to see if the mass of NaCl we calculated will give the right concentration: m(NaCl) = 1.46 g M(NaCl) = 58.44 g mol^-1 n(NaCl) = m(NaCl) ÷ M(NaCl) = 1.46 ÷ 58.44 = 0.025 mol c(NaCl) = n(NaCl) ÷ V(NaCl_(aq)) = 0.025 ÷ 0.25 = 0.10 mol L^-1 1.46 g of NaCl in a 250.00 mL volume does make a 0.10 mol L^-1 so we are confident our solution is correct.
STOP!	State the solution.	Chris the Chemist needs to weigh out 1.46 g of NaCl(s).

Can you apply this?

Join AUS-e-TUTE!

Take the exam now!

(1) There are many other ways to measure concentration of solutions in chemistry. You will find tutorials dealing with these under the heading Solutions on the AUS-e-TUTE homepage.