Molarity

21st Nov 2019 @ 11 min read

The molarity or molar concentration of a solute is a way of measurement just like mole fraction or mass fraction. It is a property of solutions, particularly homogeneous solutions. Before we dive into molarity, let us familiarize ourselves with some terms that we will use in this article.

A solution is a homogeneous mixture made up of two or more substance. We come across many substances in our everyday life which are solutions e.g. juices, air, soft drinks, alcoholic beverages, bodily fluids like urine. Seawater is one the biggest solutions.

Seawater is a solution. It consists of water (solvent) and numerous salts (solutes) dissolved in it.

A solvent is a major component in a solution. for example, water is the solvent in seawater. On the other hand, a solute is a minor component in a solution, for example, sodium salts in seawater.

Definition

The molarity of a solute in the solution is defined as the number of moles of the solute dissolved per unit volume of the solution. It is also called the molar concentration.

Formula

We can arrive at a formula using the above definition.

$c_i=\frac{n_i}{V}$

Here, c_i is the molarity of solute i, n_i is the number of moles of solute i, and V is the volume of the solution.

Units

The preferred unit of molarity is mol dm⁻³ or mol L⁻¹.

Note: dm³ (cubic decimetre) and L (litre) are equivalent i.e. 1 dm³ = 1 L.

Although mol dm⁻³ is not an SI unit, it remains a preferred unit by chemists. The SI unit of the molarity is mol m⁻³. It is an inconvenient and impractical unit for laboratory uses. Consequently, mol dm⁻³ became an accepted unit. The symbol M is traditionally used to denote mol dm⁻³. In addition, we can also use the SI prefixes with the symbol M e.g. mM (millimolar), µM (micromolar), nM (nanomolar). The table lists some of these units.

Units of molarity with SI prefixes
Unit	Abbreviation	Equivalent in M (mol L⁻¹)
megamolar	MM	10⁶
kilomolar	kM	10³
hectomolar	hM	10²
decamolar	daM	10
decimolar	dM	10⁻¹
centimolar	cM	10⁻²
millimolar	mM	10⁻³
micromolar	µM	10⁻⁶
nanomolar	nM	10⁻⁹
picomolar	pM	10⁻¹²

Explanation

The molarity is the ratio of the number of moles of a constituent to the total volume of the solution.

$\text{Molarity}=\frac{\text{Number of moles a component}}{\text{Volume of the solution}}$

In chemistry labs, the molarity is mostly expressed in mol L⁻¹. A more practical definition will be the number of moles of a solute per litre of the solution.

$\text{Molarity (M)}=\frac{\text{Number of moles a solute}}{\text{litres of the solution}}$

0.5 M of NaCl solution means 0.5 mol of NaCl in a litre of the solution. As the value of molar concentration increases, the amount of the solute per litre also increases. For example, 1.0 M of NaCl solution will have a more amount of NaCl than 0.5 M of NaCl per litre of solution. The figure below depicts the same.

Three beakers of different molar concentrations — As molar concentration increases, the amount of the solute also increases.

Solution can be a liquid, or even a gas or solid. Consider an example of sugar in water. In this example, the solution is a liquid.

Example 1: Sugar in water

Statement: Sugar easily dissolves in water. 50 g of sugar cubes are dissolved in 200 mL of a glass of water. Find the molar concentration of the sugar in the solution if the molar mass of the sugar is 342 g mol⁻¹?

Molarity of sugar in a glass of water — Sugar in a glass of water

Solution: First, we have to calculate the moles of sugar using the molar mass of sugar.

$n_\text{sugar}=\frac{m_\text{sugar}}{M_\text{sugar}}& = \frac{50}{342}\\&=0.146\,\text{mol}$

The volume of solution is 200 mL = 0.200 L. Let c_sugar be the molarity.

$c_\text{sugar}=\frac{n_\text{sugar}}{V}& = \frac{0.146}{0.200}\\&=0.73\,\text{mol}\,\text{L}^{-1}$

Therefore, the molar concentration of the sugar in the solution is 0.73 mol L⁻¹ or 0.73 M.

Let take another example, but this time solution is a gas.

Example 2: Pollutants in air

Statement: The climate change and global warming is a burning issue in the world. Increasing pollutants in air is an existential threat to our future generations. One of such pollutants is ammonia (NH₃). It is mainly emitted from agricultural wastes. For a volume of 2.0 m³, the number of moles of ammonia in air is 1.2 × 10⁻⁷ mol. Find the molar concentration of ammonia in air.

Solution: The number of moles of ammonia is 1.2 × 10⁻⁷ mol, and the volume of the air is 2.0 m³. n_NH₃ = 1.2 × 10⁻⁷ mol and V = 2.0 m³.

$c_{\text{NH}_3}=\frac{n_{\text{NH}_3}}{V}& = \frac{1.2 \times 10^{-7}}{2.0}\\&=6.0\times 10^{-8}\,\text{mol}\,\text{L}^{-1} \\&=60 \,\text{nM}$

Thus, ammonia present in the air is 60 nM.

In this example below, the solution is a solid mixture.

Example 3: Bronze

Statement: A bronze is an alloy primarily comprised of 88 % copper and 12 % tin. Find the molar concentration of copper and tin in the alloy? The molar mass of copper and tin is 63.5 g mol⁻¹ and 118.7 g mol⁻¹. Assume the density of the alloy 8.30 g L⁻¹.

Solution: Assume 100 g of bronze. The mass of copper and tin is 88 g and 12 g. We can determine their moles using respective molar masses.

$n_\text{Cu}=\frac{88}{63.5}=1.39 \,\text{g}\,\text{mol}^{-1}$ $n_\text{Sn}=\frac{12}{118.7}=0.101 \,\text{g}\,\text{mol}^{-1}$

The volume of the alloy is mass divided by density.

$V=\frac{m}{\rho}=\frac{100}{8.30} = 12.0 \,\text{L}$

Let c_Cu and c_Sn be the molar concentration of copper and tin.

$c_\text{Cu}=\frac{n_\text{Cu}}{V}=\frac{1.39}{12.0}\approx 0.12 \,\text{M}$ $c_\text{Sn}=\frac{n_\text{Sn}}{V}=\frac{0.101}{12.0} &= 0.008417 \,\text{mM} \\& \approx 8.4 \,\text{mM}$

For preparation of solutions

Every science student has experienced with solution preparation in chemical labs. Two primary inputs needs for preparing a solution are a desired molar concentration (what concentration we want?) and the volume of the solution (how much we want?).

To prepare solutions — A flask, beaker and graduated cylinder with a weighing balance

There are two methods through which we can make a solution:

To make a stock solution
From a stock solution (dilution)

Note: A stock solution is a concentrated solution that is prepared in advance, and later can be diluted to give a solution of lower concentration.

Both methods are discussed below in brief.

To make a stock solution

A stock solution is prepared directly from chemicals itself. In other words, we have to weigh or measure the solute and mix it in a solvent.

By knowing the molarity and the volume of solution, we can calculate the amount of the solute present in it. Let take an example below to explain the same.

Example 4: Sodium chloride

Statement: Sodium chloride (NaCl) is a common salt. It is an ionic compound compose of sodium ions (Na⁺) and chloride ions (Cl⁻). It is also responsible for the salinity of seawater. Let say we will be conducting an electrolysis experiment. The volume of NaCl require for the experiment is 200 mL and its concentration is 1.8 M. Find the amount of NaCl require?

Sodium chloride salt — A spoon of sodium chloride

Solution: The volume and concentration of NaCl is available. c_NaCl = 1.8 M and V = 200 mL = 0.200 L.

n_NaCl is the number of moles of NaCl.

$c_\text{NaCl}=\frac{n_\text{NaCl}}{V}$ $n_\text{NaCl}=V\times c_\text{NaCl} &= 0.200 \times 1.8 \\&=0.360\, \text{mol}$

The molar mass of NaCl is 58.5 g mol⁻¹.

$m_\text{NaCl}=n_\text{NaCl} \times M_\text{NaCl} &= 0.360 \times 58.5 \\&= 21 \text{g}$

Therefore, we have to add 21 g of NaCl in 200 mL of water to give 1.8 M of NaCl.

From a stock solution (dilution)

In dilution, a concentrated solution is diluted by pouring the solvent in it. We usually prepare solutions by dilution. It is easy and time-efficient method because we only have to dilute an existing concentrated solution.

Consider the following parameters relating a dilution:

M₁ is the initial concentration of a solution—before the dilution. This is known to us.
V₁ is the volume of the concentrated solution that we have to dilute. This is unknown to us.
M₂ is the final concentration of a solution—after the dilution. This is known to us.
V₂ is the volume of the diluted solution. This is also known to us.

In dilution, only the amount of solvent changes, the amount of a solute remains constant. Let n_solute be the number of moles of a solute.

$M_1=\frac{n_\text{solute}}{V_1} \qquad \text{and}\qquad M_2=\frac{n_\text{solute}}{V_2}$

Eliminating n_solute from the above two equations,

$M_1V_1 = M_2 V_2$

This is a very helpful equation. It is the law of conservation of mass in dilution. Consider an example to better understand it.

Example 5: Sulphuric acid

Statement: Sulphuric acid (H₂SO₄) is a common chemical reagent. It is also one of the largest produced chemical in the world. For a titration, we want 100 mL of 0.01 M H₂SO₄. 0.10 M of H₂SO₄ is available in the lab. Find the volume of water to be added in 0.10 M of H₂SO₄?

Sulphuric acid bottle
[Image source: Wikimedia]

Solution: The desired concentration and volume of H₂SO₄ is 0.01 M and 100 mL. M₂ = 0.01 M and V₂ = 100 mL. And the available concentration is 0.10 M. M₁ = 0.10 M.

Calculating V₁,

$V_1 = \frac{M_2 V_2}{M_1} &= \frac{0.01 \times 100}{0.1} \\&= 10 \,\text{mL}$

This is the volume of 0.10 M of H₂SO₄ require for the dilution. The amount of water to be added is the difference between two volumes.

$V_{\text{H}_2\text{O}} = V_2-V_1&=100-10\\&=90\,\text{mL}$

Therefore, we have to add 90 mL of water in 0.1 M H₂SO₄ to give 100 mL of 0.01 M H₂SO₄.

Ionic compounds

Ionic compounds dissociate to form ions when they are dissolved in polar solvents like water. The molar concentration of an ionic compound as a whole may differ from the molar concentration of its ions.

Example 6: Ferric chloride

Statement: Ferric chloride (FeCl₃) is an inorganic corrosive compound. Its solution is greenish in colour. A molecule of FeCl₃ contains one Fe³⁺ ion and three Cl⁻ ions.

$\ce{FeCl3-> Fe^{3+} $+$ 3Cl-}$

In a solution, the mass of FeCl₃ is 50 g. Find the molar concentration of FeCl₃, Fe³⁺, and Cl⁻ if the volume of the solution is 2.0 L?

Ferric chloride solution
[Image source: Wikimedia]

Solution: The molar mass of FeCl₃ is 162 g mol⁻¹.

$n_{\text{FeCl}_3}=\frac{50}{162}= 0.309\,\text{mol}$

For 1 mol of FeCl₃ dissociates, 1 mol of Fe³⁺ ions and 3 mol of Cl⁻ ions are formed. For 0.309 mol of FeCl₃, we have 0.309 mol of Fe³⁺ ions and 0.927 mol of Cl⁻ ions.

Calculating the molar concentration for each species,

$c_{\text{FeCl}_3}=\frac{n_{\text{FeCl}_3}}{V} &=\frac{0.309}{2.0} \\&=0.15\,\text{M}$ $c_{\text{Fe}^{3+}}=\frac{n_{\text{Fe}^{3+}}}{V} &=\frac{0.309}{2.0} \\&=0.15\,\text{M}$ $c_{\text{Cl}^{-}}=\frac{n_{\text{Cl}^{-}}}{V} &=\frac{0.927}{2.0} \\&=0.46\,\text{M}$

Here, FeCl₃ is the original species, and its molar concentration is also called as formal concentration.

Properties of molarity

The molarity is a property of a homogeneous solution.
It is intensive property since it is the ratio of two extensive properties. Hence, its value will not alter from sample to sample.
However, its value may change with temperature and pressure. The reason for this is the volume of solution is a function of temperature and pressure.

Relation with other quantities

Avogadro's constant

Avogadro's constant (N_A) is the number of molecules in a mole of substance. Let N_i be the molecules of a solute.

$n_i=\frac{N_i}{N_\text{A}}$

Substituting the above expression,

$c_i=\frac{n_i}{V}=\frac{N_i\big/N_\text{A}}{V}=\frac{N_i\big/V}{N_\text{A}}$

Here, the ratio ^N_i⁄_V is the number of molecules per unit volume, aka number density.

Mass fraction

The mass fraction of a component is the ratio of the mass of the component to the total mass of the solution.

$mass fraction$

Let M_i and ρ be the molar mass of the solute and density of the solution.

$mass fraction to molarity$

Mass concentration

The mass concentration of a solute is the ratio of mass of the solute to the volume of the solution.

$\rho_i= \frac{m_i}{V}$

Using the above formula,

$c_i=\frac{n_i}{V}&=\frac{m_i\big/M_i}{V}\\&=\frac{\rho_i}{M_i}$

Mole fraction

The molar fraction of a component is the ratio of the moles of the component to the total moles.

$mole fraction$

Let M be the average molar mass of the solution. $M=\sum x_i M_i$ .

$molarity to mole fraction$

Molality

The molality of a solute is the ratio of the number of moles of the solute to the mass of a solvent.

Consider a binary mixture. Let subscript 1 and subscript 2 denote solvent and solute.

$b_2 = \frac{n_2}{m_1}$

The molar concentration of the solute is as follows:

$molality to mass fraction$

Solved problems

Problem 1: Lead iodide precipitation

Statement: When an aqueous solution of lead nitrate (Pb(NO₃)₂) is brought in contact with aqueous potassium iodide (KI), lead iodide (PbI₂) is precipitated as a yellow solid.

$\ce{Pb(NO3)2 $+$ 2KI -> PbI2 v $+$ 2KNO3}$

Find the volume of 0.05 M of KI require for a complete precipitation of 0.20 L of 0.02 M Pb(NO₃)₂?

Solution: For every mole of Pb(NO₃)₂ consumed, 2 moles of KI is consumed. In other words, the number of moles of KI is twice the number of moles of Pb(NO₃)₂. Thus, n_KI = 2n_Pb(NO₃)₂.

But we also know n_Pb(NO₃)₂ = M_Pb(NO₃)₂V_Pb(NO₃)₂ and n_KI = M_KIV_KI. Substituting these values in the above equation,

$M_{\text{KI}} V_{\text{KI}}=2M_{\text{Pb}(\text{NO}_3)_2} V_{\text{Pb}(\text{NO}_3)_2}$ $V_{\text{KI}} &= \frac{2M_{\text{Pb}(\text{NO}_3)_2} V_{\text{Pb}(\text{NO}_3)_2}}{M_{\text{KI}}}\\&=\frac{2 \times 0.02 \times 0.20}{0.05}\\&=0.16\,\text{L}$

Therefore, 0.16 L of 0.05 KI will completely precipitate 0.02 L of 0.02 M Pb(NO₃)₂.

Problem 2: Hydrogen chloride dilution

Statement: 0.50 M of Hydrogen chloride (HCl) solution is available in the lab. For a neutralization reaction, we require 0.10 L of 0.10 M HCl. Find the volume of water to be added in 0.50 M HCl?

Solution: Let M₁ and M₂ be the molarity of 0.50 M and 0.10 M HCl.

$M_1V_1=M_2V_2$ $V_1=\frac{M_2V_2}{M_1}=\frac{0.10 \times 0.10}{0.50} = 0.02\,\text{L}$

The volume of water to be added is V₁ − V₂.

$V_{\text{H}_2\text{O}}=V_1-V_2 = 0.10-0.02=0.08\,\text{L}$

Problem 3: Sodium hydroxide

Statement: Find the volume of 5.0 M of NaOH if the number of moles of NaOH in the solution is 2.5 mol?

Solution: M_NaOH = 5.0 M and n_NaOH = 2.5 mol.

V_NaOH is the volume of NaOH.

$V_\text{NaOH}=\frac{n_\text{NaOH}}{M_\text{NaOH}}=\frac{2.5}{5.0}=0.50\,\text{L}$

Practice problems

Problem 1: Find the molar concentration of hydrogen and sulphate ions in 0.3 M sulphuric acid (H₂SO₄)?

Problem 2: Find the litres of water to be added to dilute 0.10 M of sodium hydroxide (NaOH) to 1.0 L of 0.050 M NaOH?

Problem 3: In the below precipitation reaction, the molarity of 0.20 L of silver nitrate (AgNO₃) is 0.02 M.

$\ce{AgNO3 $+$ KCl -> AgCl v + KNO3}$

Find the litres of 0.01 M KCl require for complete consumption of AgNO₃?

Summary

The molarity of a solute is the ratio of the number of moles of the solute to the volume of the solution.
Its formula is $c_i=\frac{n_i}{V}$ .
It is usually expressed in g L⁻¹ (≡ g dm⁻³). This unit is denoted by the symbol M.
$1 \,\text{M (Molar)} =\frac{1 \,\text{mol of a solute}}{1\,\text{L of the solution}}$
For dilution, $M_1V_1 = M_2V_2$ .
Molarity is an intensive property; it will not change from sample to sample, but it may change with temperature and pressure.

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