There are many ways of expressing the concentration of a solution These methods are as follows-

It may be defined as the number of parts by mass of solute per hundred parts by mass of solution.

For example, a `5%` (by mass) solution of sugar in water means that `100` g of solution contain `5` g of sugar.

Mass Percentage = `text(Mass of any component)/text(Total mass)` `xx 100`

It may be defined as the number of parts by volume of solute per hundred parts by volume of solution. For example. a `25%` (by volume) solution of ethyl alcohol means that `100 cm^3` of the solution contain `25 cm^3` of ethyl alcohol and `75 cm^3` of water.

Volume percentage = `text( Volume of component ) / text(total volume of solution)` `xx 100`

If `w_A` is the mass of component `A` and `w_B` the mass of component B in the solution, then the mass fraction of component `A` and `B` is written as

Mass fraction of `A = w_A/(w_A + w_B)`

Mass fraction of `B =w_B/(w_A+w_B)`

Mole fraction may be defined as the ratio of the number of moles of one component to the total number of moles of all the components (i.e. solute and solvent) present in the solution. Let us suppose that a solution contains the components `A` and `B` and suppose that `a` gram of `A` and `b` gram of `B` are present in it. Let the molecular masses of `A` and `B` are `M_A` and `M_B` respectively.

Then number of moles of `A` are given by `n_A = a/M_A`

and number of moles of `B` are given by `n_B = b/M_B`

Total number of moles of `A` and `B` `= n_A + n_B`

Mole fraction of `A`, `X_A = n_A/(n_A+n_B)`

Mole fraction of `B`, `X_B = n_B/(n_A+n_B)`

Sum of mole fractions of all components is always one.

i.e., `X_A +X_B = n_A/(n_A+n_B) + n_B/(n_A+n_B) =1`

So, if mole fraction of one component of a binary solution is known say `X_B`, then the mole fraction of `X_A = 1 - X_B`.

It may be noted that the mole fraction is always independent of the temperature.

Mole percent is the number of moles of a component in `100` mole of the solute and solvent.

Mole percent = Mole fraction `xx 1 00`

Example : for a solution containing `8` moles `NH_3` and `12` moles of `H_2O`.

Mole fraction of `NH_3 , X_(NH_3) = (8 text(mole))/(12 text(moles) + 8text(moles)) = 2/5`

Mole percent of `NH_3 = 2/5 xx 100 = 40 text(mole) %`

When a solute is present in trace amounts, its concentration is expressed in parts per million. It may be defined as the number of parts by mass of solute per million parts by mass of the solution.

Parts per million (ppm) = `text(Mass of solute)/text(Mass of solution)` `xx 10^6`

Molarity of a solution is defmed as the number of moles of the solute dissolved per litre of the solution. It is represented by capital `M`. Mathematically,

Molarity (M) = `text(Mass of the solution in gram per litre)/text(Molecular mass of the solute)`

A solution having molarity "one" is called molar solution.

It may be remembered that both normality as well as molarity of a solution changes with change in temperature.

Molality of a solution may be defined as the number of moles of the solute dissolved in `1000` gm (`1` kg) of the solvent. It is represented by small `m`. Mathematically

Molality (`m`) = `text(Mass of the solute in gram per kg of solvent)/text(Molecular mass of the solute)`

A solution containing one mole of solute per `1000` gm of solvent (`1` kg) has molality equal to one and is called molal solution. Molality is expressed in units of moles per kilogram (mol `kg^(-1)`)

The molality of a solution does not change with temperature.

Normality of a solution is defined as the number of gram equivalent of the solute dissolved per litre of the solution. It is represented by `N`. Mathematically,

Normality `N` = `text(Mass of solution gram per litre)/text(Equivalent mass of the solute)`

A solution having normality equal to one is called "normal solution". Such a solution contains one gram equivalent of solute per litre of solution. A seminormal solution contains `0.5` gram equivalent of solute. A decinormal solution contains `0.1` gram equivalent and a centinormal solution contains `0.01` gram equivalent of solute per litre of solution.

or Normality = `(text[Mass of the solution]xx 1000)/(text[Equivalent mass of the solute]xx V)`

Where `V` is the volume in milliliter.

It may be defined as the number of gram formula masses of the ionic solutes dissolved per litre of the solution. Mathematically,

Formality (`F`) = `text(Mass of the ionic in gram per litr)/text(Formula mass of the solute)`

Commonly, the term formality is used to express the concentration of the ionic solids which do not exist as molecule but as network of ions.