Molarity Formula – Definition of Mole & Calculation

Molarity Formula: In chemistry, the concentration of solutions plays a vital role in predicting the nature of a particular chemical reaction. This is because the concentration of solutions accounts for the number of particles participating in bringing about a chemical reaction. Measurements based on either mass or volume are not helpful for chemical reactions. A measurement unit based on moles is preferable to determine the concentration of solutions. The two most commonly used concentration units are Molarity and molality. Read on to know their definitions, formula and SI units.

What is a mole?

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 elemental entity refers to atoms, ions or molecules. The value \(6.02214076 \times {10^{23}}\) is known as Avogadro’s number represented by \({{\rm{N}}_{\rm{A}}}{\rm{.}}\) In other words, a mole is the amount of substance that contains as many entities (atoms or other particles) as there are atoms in \(12\) grams of pure carbon\(-12.\)

Mathematically, it is expressed as–

\({\rm{Moles = }}\frac{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{substance }}}}{{{\rm{ Molar}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{substance }}}}\)

Based on moles, there are two concentration units. These are – Molarity and Molality

Molarity

The Molarity \(\left( {\rm{M}} \right)\) of a solution is defined as the number of moles of a solute dissolved in one litre of the solution.

Mathematically, it is expressed as–

\({\rm{Molarity(M) = }}\frac{{{\rm{ Moles}}\,{\rm{of}}\,{\rm{Solute }}}}{{{\rm{ Volume}}\,{\rm{of}}\,{\rm{Solution }}}}\)

Its unit is \({\rm{mol/L}}\)

Molarity depends on the volume of a solution expressed in litres and not the volume of solvent. The Molarity of a solution is denoted by \({\rm{M}}\) and is read as “molar”. For example, a solution labelled as \(1.5\,{\rm{M}}\,{\rm{N}}{{\rm{H}}_3}\) is read as “\(1.5\) molar ammonia solution”.

The Molarity of a solution is dependent on the changes in physical properties of the system such as pressure and temperature. The Molarity of a solution is said to be one molar when one mole of solute is dissolved in a litre of solution.

Example:

Calculate the Molarity of a solution prepared by dissolving \(42.23\;{\rm{g}}\) of \({\rm{N}}{{\rm{H}}_4}{\rm{Cl}}\) in water to make \(500.0\;{\rm{mL}}\) solution. 

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 \( = 500{\rm{ml}} = 0.5\;{\rm{L}}\)

\({\rm{Number}}{\mkern 1mu} {\rm{of}}{\mkern 1mu} {\rm{moles}} = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{N}}{{\rm{H}}_{\rm{L}}}{\rm{Cl}}}}{{{\rm{Molar}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{N}}{{\rm{H}}_4}{\rm{Cl}}}} = \frac{{42.23}}{{53.50}} = 0.7893\;{\rm{mol}}\)

\({\rm{Molarity}}\left( {\rm{M}} \right){\rm{ = }}\frac{{{\rm{Moles}}\,{\rm{of}}\,{\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{Cl}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{the}}\,{\rm{solution}}\,{\rm{in}}\,{\rm{L}}}}{\rm{ = }}\frac{{{\rm{0}}.{\rm{7893}}\;{\rm{mol}}}}{{{\rm{0}}.{\rm{5}}\;{\rm{L}}}}{\rm{ = 1}}.{\rm{579}}\,{\rm{M}}\)

Molarity Equation

In the laboratory, chemists frequently prepare solutions of known Molarity. The primary task is to calculate the mass of the solute that is necessary to dissolve in a solvent. 

To calculate the volume of an unknown solution from a solution of known Molarity, the following equation is used:

\({{\rm{M}}_1}{{\rm{V}}_1}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} {{\rm{M}}_2}{{\rm{V}}_2}\)

Where \({{\rm{M}}_{\rm{1}}} = \) Initial Molarity of the given solution.
\({{\rm{M}}_{\rm{2}}} = \) Molarity of the new solution.
\({{\rm{V}}_{\rm{1}}} = \) Initial volume of the given solution.
\({{\rm{V}}_{\rm{2}}} = \) volume of the new solution

Suppose we have three solutions of the same solute with Molarity \({{\rm{M}}_{\rm{1}}},\,{{\rm{M}}_{\rm{2}}},\,{{\rm{M}}_{\rm{3}}}\) and volume \({{\rm{V}}_{\rm{1}}}{\rm{,}}{{\rm{V}}_{\rm{2}}}{\rm{,}}{{\rm{V}}_{\rm{3}}}.\)

Molarity of the resulting solution formed by mixing of these solution, is given by-

\({{\rm{M}}_{\rm{R}}}{\rm{ = }}\frac{{{{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{\rm{ + }}{{\rm{M}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}{\rm{ + }}{{\rm{M}}_{\rm{3}}}{{\rm{V}}_{\rm{3}}}{\rm{ + }}…}}{{{{\rm{V}}_{\rm{1}}}{\rm{ + }}{{\rm{V}}_{\rm{2}}}{\rm{ + V}}_{\rm{3}}^{\rm{ + }} \ldots }}\)

The above equation represents a concept which means the amount of moles in the solution remains constant irrespective of the change in the concentration of the solution or the volume of the solution.

Calculation of the Molarity of a Solution

Step 1. Calculation of the molar mass of the solute. 

The molar mass of the solute can be calculated using the molar mass of the elements.
Consider the example of \({\rm{NaOH:}}\)
Molar mass of \({\rm{NaOH = }}\) Molar mass of \({\rm{Na+}}\) Molar mass Of \({\rm{O+}}\) Molar mass of \({\rm{H}}\)
\({\rm{ = 23 + 16 + 1 = 40\;g/mol}}\)

In \({{\rm{H}}_{\rm{2}}}{\rm{O,}}\) molar mass of water \({\rm{ = 2}} \times \) molar mass of \({\rm{H+}}\) Molar mass of \({\rm{O}}\)

\({\rm{ = 2 + 16 = 18\;g/mol}}\)

Step 2. Calculation of the Moles of the solute.

The moles of a solute is given by,

\({\rm{n = }}\frac{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{solute}}\,{\rm{(g)}}}}{{{\rm{ Molar}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{solute}}\,{\rm{(g)}}}}\)

Step 3. Calculation of the Volume of Solution.

The volume of the solution is calculated by using the following relations:

\(1\;{\rm{c}}{{\rm{m}}^3} = 1\;{\rm{mL}} = 1000\;{\rm{m}}{{\rm{m}}^3}\)

\({\rm{1Litre = 1000\;mL = 1000\;c}}{{\rm{m}}^{\rm{3}}}{\rm{ = 1d}}{{\rm{m}}^{\rm{3}}}\)

\({\rm{1}}{{\rm{m}}^{\rm{3}}}{\rm{ = 1}}{{\rm{0}}^{\rm{6}}}{\rm{\;c}}{{\rm{m}}^{\rm{3}}}{\rm{ = 1000L}}\)

Step 4. Calculation of Molarity by using the Molarity Formula
The Molarity of the solution is calculated by using moles of the solute (step \(2\)) and the volume of the solution (step \(3\))

\({\rm{Molarity(M) = }}\frac{{{\rm{ Moles}}\,{\rm{of}}\,{\rm{Solute }}}}{{{\rm{ Volume}}\,{\rm{of}}\,{\rm{Solution}}\,{\rm{(in}}\,{\rm{L)}}}}\)

Molar Mass Formula

In chemical reactions, it is important to consider the number of atoms of each element present in each sample. Even a small quantity of a substance will contain millions of atoms, so chemists generally use the mole as the unit for the amount of substance.

amu vs. \({\rm{g/mol}}\)

Each mole of a given pure substance has a definite mass. The mass of one mole of atoms of a pure element in grams is equivalent to the atomic mass of that element in atomic mass units \(\left( {{\rm{amu}}} \right)\) or in grams per mole \(\left( {\frac{{\rm{g}}}{{{\rm{mol}}}}} \right){\rm{.}}\) \({\rm{g/mol}}\) is the most useful system of units for laboratory chemistry.

Calculation of Molar Mass of a Substance

Molar mass is the mass of a given substance divided by the amount of that substance. The molar mass is an intensive property of the substance that doesn’t depend upon the dimensions of the sample.

\({\rm{Molar}}\,{\rm{mass = }}\frac{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{given}}\,{\rm{substance }}}}{{{\rm{ Amount}}\,{\rm{of}}\,{\rm{substance }}}}\)

It is measured in \(\frac{{\rm{g}}}{{{\rm{mol}}}}{\rm{.}}\) The molar mass of an element is simply the atomic mass in gmol.

For example, the atomic mass of sodium \(\left( {{\rm{Na}}} \right)\) is \({\rm{22}}{\rm{.98}}\,{\rm{amu}}\) or \( 22.98\frac{{\rm{g}}}{{{\rm{mol}}}}.\) This means that in \({\rm{22}}{\rm{.98}}\,{\rm{grams}}\,\) of sodium, there is one mole or \(6.02214076 \times {10^{23}}\) Sodium atoms.

The molar mass of a compound can be calculated by adding up all the atomic mass of the constituent atoms.

For example, the molar mass of \({\rm{NaCl}}\) can be calculated as follows:

Atomic mass of sodium \(\left( {{\rm{22}}{\rm{.98}}\,{\rm{g/mol}}} \right){\rm{ + }}\) Atomic mass of chlorine \(\left( {{\rm{35}}{\rm{.5}}\,{\rm{g/mol}}} \right) = 58.44\,{\rm{g/mol}}{\rm{.}}\)

Learn Concept of Mole

Difference between Molarity and Molality Formula

Molality: The molality \(\left( {\rm{m}} \right)\) of a solution is defined as the number of moles of the solute present per kilogram of the solvent.

Mathematically, it is expressed as–

\({\rm{Molarity}}\,\left( {\rm{m}} \right){\rm{ = }}\frac{{{\rm{Moles}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Kg}}\,{\rm{of}}\,{\rm{Solvent}}}}\)

Its unit is \(1.0\,{\rm{mol}}.\) Molality is represented by m and is also known as ‘molal solution’. A solution that contains \(1.0\,{\rm{mol}}\) of glucose dissolved into \(1.0\,{\rm{kg}}\) of water is a “one-molal” solution of glucose. As molality is not dependent on the volume of the solution, it is independent of pressure and temperature changes. This means molality does not change with a change in pressure and temperature of the system.

Example:

A solution is prepared by dissolving \(29.1\;{\rm{g}}\) of toluene in \(832\;{\rm{g}}\) of benzene. Calculate its molality.

Mass of toluene given \( = 29.1\;{\rm{g}}\)

Molar mass of toluene \( = 92\;{\rm{g}}/{\rm{mol}}\)

Mass of Solvent \( = 832\;{\rm{g}} = 0.832\;{\rm{kg}}\)

\({\rm{Number}}\,{\rm{of}}\,{\rm{moles}} = \frac{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{toluene }}}}{{{\rm{ Molar}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{toluene }}}} = \frac{{29.1}}{{92}} = 0.316\;{\rm{mol}}\)

\({\rm{Molality(m)}} = \frac{{{\rm{ Moles}}\,{\rm{of}}\,{\rm{toluene }}}}{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{solvent}}\,{\rm{in}}\,{\rm{kg}}}} = \frac{{0.316\;{\rm{mol}}}}{{0.832\;{\rm{kg}}}} = 0.380\;{\rm{m}}\)

Summary

We deal with various types of solutions in our day to day life. Hence, it is important to know their concentrations. Molarity is an important unit to express the concentration of solutions. The volume of the solution plays a vital role in expressing the concentration of a solution in Molarity. In this article, we learnt molarity, molality, their units and ways to calculate them. We also learnt about molar mass, without which the Molarity and molality of a solution cannot be calculated.

Frequently Asked Questions (FAQs) on Molarity Formula

Q.1. How do we calculate Molarity?
Ans: The Molarity of a solution is calculated by the following formula:
\({\rm{Molarity(M) = }}\frac{{{\rm{ Moles}}\,{\rm{of}}\,{\rm{Solute }}}}{{{\rm{ Volume}}\,{\rm{of}}\,{\rm{Solution}}\,{\rm{in}}\,{\rm{Litre }}}}\)

Q.2. What is Molarity, and write its formula?
Ans: The Molarity \({\rm{(M)}}\) of a solution is defined as the number of moles of a solute dissolved in one litre of the solution.
Mathematically, it is expressed as:
\({\rm{Molarity(M) = }}\frac{{{\rm{ Moles}}\,{\rm{of}}\,{\rm{Solute }}}}{{{\rm{ Volume}}\,{\rm{of}}\,{\rm{Solution}}\,\left( {{\rm{in}}\,{\rm{L}}} \right){\rm{ }}}}\)  

Q.3. What is the Molarity of \({\rm{HCl}}\)?
Ans: Molarity is defined as the ratio of moles of solute to the volume of solution. By dividing the number of moles of \({\rm{HCl}}\) by the volume \(\left( {\rm{L}} \right)\) of the solution in which it was dissolved, we will obtain the molarity of the \({\rm{HCl}}\) solution.

Q.4. How is the Molar Mass of a substance calculated?
Ans: The molar mass of a substance is calculated by the given formula
\({\rm{Molar}}\,{\rm{mass = }}\frac{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{given}}\,{\rm{substance }}}}{{{\rm{ Amount}}\,{\rm{of}}\,{\rm{the}}\,{\rm{substance }}}}\)

Q.5. What is the difference between Molarity and Molality?
Ans: The difference between Molarity and Molality is summarised in the table given below:

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