Units of Concentration: Definition, Molarity, Normality

We come across the term “concentration” in everyday life. The unit of concentration of a solution can be defined as when a mixture of one or more solutes is dissolved in a solvent. For example; our everyday coffee, tea. The solvent makes up the major portion of the solution, whereas solute forms the minor part.

In chemistry, the concentration of the solution is defined as the amount of solute in a solvent. When a solution has more solute, it is referred to as a concentrated solution and the solution containing more solvent is referred to as a dilute solution. This article will discuss various units of concentration for dilute and concentrated solutions and the normality and molarity of a solution with examples.

What is Solubility?

In our daily life, we come across various solutions being diluted or concentrated.
A dilute solution means the quantity of solute is relatively very small, and a concentrated solution implies the solution has a large amount of solute. Tagging a solution as dilute or concentrated is a very general way of representing the concentration of solutions. However, these terms are relative and do not give us any quantitative idea of the solution.

Relative Concentration Units

Concentrations are often expressed in terms of relative units (e.g. percentages). The commonly used percentage concentrations are-

Mass Percent

The mass percent is used to express the concentration of a solution when the mass of a solute and the mass of a solution are given. Mathematically, it is expressed as:
\({\rm{Mass}}{\mkern 1mu} \,\% = \frac{{{\rm{Mass}}{\mkern 1mu} \,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Mass}}{\mkern 1mu} \,{\rm{of}}\,{\rm{solution}}}} \times 100{\mkern 1mu} \% \)
\({\rm{Mass}}\left( {\frac{{\rm{w}}}{{\rm{W}}}} \right)\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Mass}}\,{\rm{of}}\,{\rm{solute}}\,{\rm{ + }}\,{\rm{Mass}}\,{\rm{of}}\,{\rm{solvent}}}} \times 100\% \)

Example – Calculate the mass percent of \(5\,{\rm{g}}\) of caustic soda dissolved in \(100\,{\rm{g}}\) of water?
\({\rm{The}}\,{\rm{total}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{solution}} = {\rm{mass}}\,{\rm{of}}\,{\rm{caustic}}\,{\rm{soda}} + {\rm{mass}}\,{\rm{of}}\,{\rm{water}}\)
\( = 5 + 100 = 105\,{\rm{g}}\)
\({\rm{Mass}}\,\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Mass}}\,{\rm{of}}\,{\rm{solute}} + {\rm{Mass}}\,{\rm{of}}\,{\rm{solvent}}}} \times 100\% \)
\({\rm{Mass}}\,\% = \frac{5}{{5 + 100}} \times 100\% = \frac{5}{{105}} \times 100\% = 4.761\% \)
Therefore, the mass percent of \(5\,{\rm{g}}\) of caustic soda dissolved in \(100\,{\rm{g}}\) of water is \(4.761\% .\)

Volume Percent

The volume percent is used to express the concentration of a solution when the volume of a solute and the volume of a solution is given. Mathematically, it is expressed as:
\({\rm{Volume}}\,\% = \frac{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solution}}}} \times 100\% \) of
\({\rm{Volume}}{\mkern 1mu} \left( {\frac{{\rm{v}}}{{\rm{v}}}} \right)\% = \frac{{{\rm{Volume}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} \,{\rm{Solute}}}}{{{\rm{Volume}}{\mkern 1mu} \,{\rm{of}}\,{\rm{Solute}} + {\rm{Volume}}{\mkern 1mu} \,{\rm{of}}\,{\rm{Solvent}}}} \times 100\% \)
Example – Rubbing alcohol is a \(70\% \,\left( {\frac{{\rm{v}}}{{\rm{v}}}} \right)\) solution. This means \(70\,{\rm{ml}}\) of Rubbing alcohol is dissolved in \(30\,{\rm{ml}}\) of water.

Mass/Volume Percent

The \(\frac{{{\rm{mass}}}}{{{\rm{volume}}}}\) per cent is used to express the concentration of a solution when the mass of the solute and volume of the solution is given. It measures the mass or weight of solute in grams (e.g., in grams) vs the volume of solution (e.g., in \({\rm{mL}}\)). Though the numerator and denominator have different units, this concentration unit is not a true relative unit (e.g., percentage); however, it is often expressed in percentage. Mathematically, it is expressed as:

\(\frac{{{\rm{Mass}}}}{{{\rm{Volume}}}}\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solution}}}} \times 100\% \)
\(\frac{{{\rm{Mass}}}}{{{\rm{Volume}}}}\left( {\frac{{\rm{m}}}{{\rm{v}}}} \right)\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}} + {\rm{Volume}}\,{\rm{of}}\,{\rm{Solvent}}}} \times 100\% \)

Example – A pharmacist adds \(2.00\,{\rm{ml}}\) of distilled water to \(4.00\,{\rm{g}}\) of a powdered drug. The final volume of the solution is \(3.00\,{\rm{ml}}.\) What is the \(\frac{m}{V}\) of the solution?
\(\frac{{{\rm{Mass}}}}{{{\rm{Volume}}}}\left( {\frac{{\rm{m}}}{{\rm{v}}}} \right)\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}} + {\rm{Volume}}\,{\rm{of}}\,{\rm{Solvent}}}} \times 100\% \)
\(\frac{{{\rm{Mass}}}}{{{\rm{Volume}}}}\left( {\frac{{\rm{m}}}{{\rm{v}}}} \right)\% = \frac{4}{3} \times 100\% = 133\% \)
This means that the concentration is \(\frac{{{\rm{133}}\,{\rm{g}}}}{{{\rm{100}}\,{\rm{ml}}}}.\)

Dilute Concentrations Units

Sometimes when solutions are too dilute, it is inconvenient to express them in percentages such as \(0.00001\% \) or \(0.000000001\% .\) Such low concentrations are expressed efficiently in a different set of units. These are-
\(1\,{\rm{g}}\) of solute dissolved in \(1000000\,{\rm{mL}}\) of a solution creates a very small percentage concentration (same as \(1\,{\rm{mg}}\) solute dissolved in \(1\,{\rm{L}}\) solution). Because a solution like this would be so dilute, it is better expressed in the following ways-

Parts per Million

A part per million is one part of solute per million parts of the solution. In terms of defining equations, we write as:
\({\rm{ppm}}\left( {\frac{{\rm{w}}}{{\rm{W}}}} \right)\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}\,{\rm{ + }}\,{\rm{Mass}}\,{\rm{of}}\,{\rm{Solvent}}}} \times {10^6}\)
\({\rm{ppm}}\left( {\frac{{\rm{v}}}{{\rm{V}}}} \right)\% = \frac{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}}\,{\rm{ + }}\,{\rm{Volume}}\,{\rm{of}}\,{\rm{Solvent}}}} \times {10^6}\)
\({\rm{ppm}}\left( {\frac{{\rm{m}}}{{\rm{V}}}} \right)\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}}\,{\rm{ + }}\,{\rm{Volume}}\,{\rm{of}}\,{\rm{Solvent}}}} \times {10^6}\)

Parts per Billion

Parts per billion \(\left( {{\rm{ppb}}} \right)\) is similar to \({\rm{ppm,}}\) except \(1\,{\rm{ppb}}\) is \(1000\) times more dilute than \(1\,{\rm{ppm}}{\rm{.}}\) A part per billion is one part of the solute per billion parts of the solution. In terms of defining equations, we write as:

\({\rm{ppm}}\left( {\frac{{\rm{w}}}{{\rm{W}}}} \right)\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}} + {\rm{Mass}}\,{\rm{of}}\,{\rm{Solvent}}}} \times {10^9}\)
\({\rm{ppm}}\left( {\frac{{\rm{v}}}{{\rm{V}}}} \right)\% = \frac{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}} + {\rm{Volume}}\,{\rm{of}}\,{\rm{Solvent}}}} \times {10^9}\)
\({\rm{ppm}}\left( {\frac{{\rm{m}}}{{\rm{V}}}} \right)\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}} + {\rm{Volume}}\,{\rm{of}}\,{\rm{Solvent}}}} \times {10^9}\)

Parts per Trillion

\(1\,{\rm{ppt}}\) is \(1000\) times more dilute than \(1\,{\rm{ppb}}\) and \(1000000\) times more dilute than \(1\,{\rm{ppm}}{\rm{.}}\)
\({\rm{ppm}}\,\left( {\frac{{\rm{w}}}{{\rm{W}}}} \right)\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}} + {\rm{Mass}}\,{\rm{of}}\,{\rm{Solvent}}}} \times {10^{12}}\)
\({\rm{ppm}}\,\left( {\frac{{\rm{v}}}{{\rm{V}}}} \right)\% = \frac{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}} + {\rm{Volume}}\,{\rm{of}}\,{\rm{Solvent}}}} \times {10^{12}}\)
\({\rm{ppm}}\,\left( {\frac{{\rm{m}}}{{\rm{V}}}} \right)\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{Solute}} + {\rm{Volume}}\,{\rm{of}}\,{\rm{Solvent}}}} \times {10^{12}}\)

Concentration Units Based on Moles

Chemists primarily need the concentration of solutions to be expressed in a way that accounts for the number of particles that react according to a particular chemical equation. Since percentage measurements are based on either mass or volume, they are generally not useful for chemical reactions.

A mole is simply a unit of measurement. A mole is defined as the amount of a substance that contains exactly \(6.02214076 \times {10^{23}}\) elementary entities’ of the given substance. The term elementary entity refers to atoms, ions, or molecules. The value \(6.02214076 \times {10^{23}}\) is known as Avogadro’s number represented by \({{\rm{N}}_{\rm{A}}}{\rm{.}}\)
Mathematically, it is expressed as:
\({\rm{Moles}} = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{substance}}}}{{{\rm{Molar}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{substance}}}}\)

Molarity

The molarity \(\left( {\rm{M}} \right)\) of a solution is used to represent the number of moles of the solute dissolved in one litre of the solution.
Mathematically, it is expressed as:
\({\rm{Molarity}}\,\left( {\rm{M}} \right) = \frac{{{\rm{Moles}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Liters}}\,{\rm{of}}\,{\rm{Solution}}}}\)
Its unit is \({\rm{mol/L}}{\rm{.}}\)

Example-
A solution is prepared by dissolving \({\rm{42}}{\rm{.23}}\,{\rm{g}}\) of \({\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{Cl}}\) into enough water to make \({\rm{500}}{\rm{.0}}\,{\rm{mL}}\) of solution. Calculate its molarity.
Molar mass of \({\rm{N}}{{\rm{H}}_4}{\rm{Cl}} = 53.50\,{\rm{g}}\)
Mass of \({\rm{N}}{{\rm{H}}_4}{\rm{Cl}}\) given \( = 42.23\,{\rm{g}}\)
Volume of the solution \( = 500\,{\rm{ml}} = 0.5\,{\rm{L}}\)
\({\rm{Number}}{\mkern 1mu} {\rm{of}}{\mkern 1mu} {\rm{moles}} = \frac{{{\rm{Mass}}\,{\mkern 1mu} {\rm{of}}\,{\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{Cl}}}}{{{\rm{Molar}}\,{\rm{mass}}{\mkern 1mu} \,{\rm{of}}\,{\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{Cl}}}} = \frac{{)2.23}}{{53.50}} = 0.7893\,{\rm{mol}}\)
\({\rm{Molarity}}{\mkern 1mu} \left( {\rm{M}} \right) = \frac{{{\rm{MOles}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} {\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{Cl}}}}{{{\rm{Volume}}{\mkern 1mu} \,{\rm{of}}\,{\rm{the}}{\mkern 1mu} \,{\rm{solution}}{\mkern 1mu} \,{\rm{in}}{\mkern 1mu} {\rm{L}}}} = \frac{{0.7893\,{\rm{mol}}}}{{0.5\,{\rm{L}}}} = 1.579\,{\rm{M}}\)

Mole Fraction

In a solution, mole fraction is defined as the ratio of moles of one component to the total moles present in the solution. It is denoted by \({\rm{X}}{\rm{.}}\) Suppose we have a solution containing \({\rm{A}}\) as a solute and \({\rm{B}}\) as the solvent. Let \({{\rm{n}}_{\rm{A}}}\) be the number of moles of \({\rm{A}}\) and \({{\rm{n}}_{\rm{B}}}\) be the number of moles of \({\rm{B}}{\rm{.}}\)

Then, Mole fraction can be expressed as:
\({{\rm{X}}_{{\rm{solute}}}} = \frac{{{\rm{Moles}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Moles}}\,{\rm{of}}\,{\rm{solute}} + {\rm{Moles}}\,{\rm{of}}\,{\rm{solvent}}}} = \frac{{{{\rm{n}}_{\rm{A}}}}}{{{{\rm{n}}_{\rm{A}}} + {{\rm{n}}_{\rm{B}}}}}\)
\({{\rm{X}}_{{\rm{solvent}}}} = \frac{{{\rm{Moles}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Moles}}\,{\rm{of}}\,{\rm{solute}} + {\rm{Moles}}\,{\rm{of}}\,{\rm{solvent}}}} = \frac{{{{\rm{n}}_{\rm{B}}}}}{{{{\rm{n}}_{\rm{A}}} + {{\rm{n}}_{\rm{B}}}}}\)
\({{\rm{X}}_{{\rm{solute}}}} + {{\rm{X}}_{{\rm{solvent}}}} = 1\)

Example: Calculate the mole fraction of a solution made by \(37.5\% \,{\rm{HCI}}\) and \(62.5\% \) of water.
\({\rm{HCI}} = 37.5\% ,\) molar mass of \({\rm{HCI}} = 36.5\,{\rm{g}}\)
Moles of \({\rm{HCI,}}\,{{\rm{n}}_{{\rm{HCI}}}} = \frac{{37.5}}{{36.5}} = 1.03\,{\rm{mol}}\)
\({{\rm{H}}_{\rm{2}}}{\rm{O}} = 62.5\% ,\) molar mass of \({{\rm{H}}_{\rm{2}}}{\rm{O}} = 18\,{\rm{g}}\)

Moles of \({{\rm{H}}_{\rm{2}}}{\rm{O,}}\,{{\rm{n}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = \frac{{62.5}}{{18}} = 3.47\,{\rm{mol}}\)
\({{\rm{X}}_{{\rm{HCl}}}} = \frac{{{{\rm{n}}_{{\rm{HCl}}}}}}{{{{\rm{n}}_{{\rm{HCl}}}} + {{\rm{n}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}} = \frac{{1.03}}{{1.03 + 3.47}} = 0.229\,{\rm{HCl}}\)
\({{\rm{X}}_{{{\rm{H}}_2}{\rm{O}}}} = \frac{{{{\rm{n}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}}{{{{\rm{n}}_{{\rm{HCl}}}} + {{\rm{n}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}} = \frac{{3.47}}{{1.03 + 3.47}} = 0.771\,{{\rm{H}}_{\rm{2}}}{\rm{O}}\)
\({{\rm{X}}_{{\rm{HCl}}}} + {{\rm{X}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = 1\)

Mole Percent

When the mole fraction is multiplied by \(100,\) it gives the mole percentage. It is also referred to as amount/amount percent (abbreviated as \(\frac{{\rm{n}}}{{\rm{n}}}\% \)). In general chemistry, all the mole percent of a mixture add up to \(100\) mole percent. Therefore, we can easily convert mole percent back to mole fraction by dividing by \(100.\)
\({\rm{Mole}}\,\left( {\frac{{\rm{n}}}{{\rm{N}}}} \right)\% = {\rm{Mole}}\,{\rm{Fraction}} \times 100\% \)
\({\rm{Mole}}\,\left( {\frac{{\rm{n}}}{{\rm{N}}}} \right)\% = \frac{{{\rm{Moles}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Moles}}\,{\rm{of}}\,{\rm{solute + Moles}}\,{\rm{of}}\,{\rm{Solvent}}}} \times 100\% \)

Normality

When \(1\,{\rm{gm}}\) equivalent weight of the solute is dissolved in \(1000\,{\rm{ml}}\) of the solution, it’s called \(1\) normal or \(1\,{\rm{N}}\) solution. It is represented by \({\rm{N}}.\)
Mathematically, it is expressed as:
\({\rm{Normality}}\,\left( {\rm{N}} \right) = \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{gram}}\,{\rm{equivalent}}\,{\rm{weight}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{the}}\,{\rm{solution}}\,{\rm{in}}\,{\rm{Litres}}}}\)
\({\rm{Number}}\,{\rm{of}}\,{\rm{gram}}\,{\rm{equivalents}} = \frac{{{\rm{Weight}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Equivalent}}\,{\rm{weight}}\,{\rm{of}}\,{\rm{solute}}}}\)
\({\rm{N}} = \frac{{{\rm{Molarity}} \times {\rm{Molar}}\,{\rm{mass}}}}{{{\rm{Equivalent}}\,{\rm{mass}}}}\)
\({\rm{N}} = {\rm{Molarity}} \times {\rm{Basically}} = {\rm{Molarity}} \times {\rm{Acidity}}\)
\({\rm{N}} = {\rm{Molarity}} \times {\rm{Valency}}\)
\({\rm{N}} = {\rm{Molarity}} \times {\rm{Number}}\,{\rm{of}}\,{{\rm{H}}^{\rm{ + }}}\) and \({\rm{O}}{{\rm{H}}^ – }\)
Its unit is eq \({{\rm{L}}^{ – 1}}.\)

Difference between Molarity, Molality, and Normality

FAQs

Q.1. What is the unit of \(\frac{{\rm{W}}}{{\rm{W}}}?\)
Ans: \(\frac{{\rm{W}}}{{\rm{W}}}\) is unitless. It represents a mass percent. The numerator and denominator of mass percent consist of identical units. Hence \(\frac{{\rm{W}}}{{\rm{W}}}\) is unitless.

Q.2. What is the unit of concentration of a solution?
Ans: The unit of concentration of a solution is a tool to measure the amount of a solute dissolved in a solution. The units of concentration are a more precise way of expressing the concentration of solutions quantitatively.

Q.3. What are the different units of concentration?
Ans: The different concentration units are molarity, molality, normality, mass percent, volume percent, mole fraction, mole percent, \(\frac{{{\rm{mass}}}}{{{\rm{volume}}}}\) percent, \({\rm{ppm,}}\,{\rm{ppb}}\) and \({\rm{ppt}}{\rm{.}}\)

Q.4. What are the units of concentration in beer’s law?
Ans: It has units of \({{\rm{M}}^{ – 1}}\,{\rm{c}}{{\rm{m}}^{ – 1}}\,\left( {{\rm{M}} = {\rm{molarity}}} \right).\) The variation of \(\varepsilon \) with wavelength is characteristic of a substance.

Q.5. What is a concentration gradient?
Ans: A concentration gradient is the rate of change of concentration with distance. It results from the unequal distribution of particles, e.g., ions, between two solutions. This imbalance of solutes between the two solutions drives solutes to move from a highly dense area to a lesser dense area. The phenomenon that exhibits concentration gradient is osmosis, diffusion, filtration etc. Its unit is \({\rm{moles/L/m}}{\rm{.}}\) 

Q.6. What are the units of concentration in molarity?
Ans: Molarity is defined as the number of moles of a solute dissolved in a litre of the solution. It depends upon temperature, pressure, and the solubility of the solute. It’s unit \({\rm{mol}}\,{{\rm{L}}^{ – 1}}.\)

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