You know what is concentration, right? But, do you know how to express it? When there is too much salt in your lemonade, you say that it is “too salty”. But in chemistry, there is nothing like “too”. We have to define it. That is why we are going to study all about the concentration of solutions in this chapter. We will look at the various methods of calculating the concentration of solutions. So, let us start!

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## Concentration of Solutions

We can calculate the concentration of solutions by various methods. Let’s study each method and determine the formulas for this method.

**1) Mass/Weight Percentage or PercentageÂ by Mass/WeightÂ **

It is the amount of solute in grams present in 100 grams of the solution. Therefore, the formula will be:

The ratio mass of solute to the mass of the solvent is theÂ mass fraction. Thus, the mass percentage of solute = Mass fraction Ã— 100. 10% solution of sugar by mass means that 10 grams of sugar is present in 100 grams of the solution, i.e., we have dissolved 10 grams of sugar in 90 grams of water.

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**Browse more Topics under Solutions**

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- Ideal and Non-ideal Solutions
- Osmosis and Osmotic Pressure
- Solubility
- Types of Solutions
- Vapour Pressure of Liquid Solutions

**2) Volume Percentage**

It is the volume of solute in mL present in 100 mL solution. The formula will be:

10% solution of HCl by volume means that 10 mL of liquid HCl is present in 100 mL of the solution.

**3) Mass by Volume PercentageÂ **

It is the mass of solute present in 100 mL of solution. We can calculate the mass of the solute using the volume percentage. The formula would be:

A 10% mass by volume solution means that 10 gm solute is present in 100 mL of solution.

**4) Molarity**

The molarity of a solution gives the number of gram molecules of the solute present in one litre of the solution.

For example, 1 mol L^{-1}Â solution of KCl means that 1 mol of KCl is dissolved in 1 L of water.Â Unit of molarity: mol L^{-1}

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**5) Molality**

Molality of a solution is the number of moles of solute dissolved in 1 Kg of the solvent.

Thus, if one gram molecule of a solute is present in 1 kg of the solvent, the concentration of solutions is said to beÂ one molal. The unit of molarity is mol kg^{-1}. MolalityÂ is the most convenient method to express the concentration of solutions because it involves the mass of liquids rather than their volumes. It is also independent of the variation in temperature.

**6) Normality Â Â **

The normality of a solution gives the number of gram equivalents of the solute present in one litre of the solution.

Thus, if one gram equivalent of a solute is present in one litre of the solution,Â the concentration of solutions is said to be 1 normal.

- 1N = NormalÂ = One gram equivalent of the solute per litre of solution Â Â = Normality is 1
- N/2Â = Seminormal Â Â = 0.5 g equivalent of the solute per litre of solution Â Â = Normality is 0.5
- N/10Â = Decinormal Â Â = 0.1 g equivalent of the solute per litre of solution Â Â = Normality is 0.1
- N/100Â = Centinormal Â = 0.01 g equivalent of the solute per litre of solution Â Â = Normality is 0.01
- N/1000Â = Millinormal Â = 0.001 g equivalent of the solute per litre of solution Â Â = Normality is 0.001

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**7) Mole FractionÂ **

The mole fraction of any component in a solution is the ratio of the number of moles of that component to the total number of moles of all components. The total mole fraction of all the components of any solution is 1. For a binary solution of A and B

Mole Fraction of A (X_{A}) =Â Â

Mole Fraction of B (X_{B})Â =Â Â

And, X_{A}+X_{B} = 1

**8) Parts per million (ppm)**

When a solute is present in trace quantities, itÂ is convenient to express the concentration of solutions in parts per million (ppm). The formula is as follows:

In case of mass, we may express it as: (Mass of solute/Mass of solution )Ã— 106

In case of volume, we may express it as: (Volume of solute/Volume of solution) Ã— 106

So, we can express the concentration of solutions in parts per million as mass to mass, volume to volume andÂ mass to volume form. Atmospheric pollution in cities is also expressed in ppm by volume. It refers to the volume of the pollutant in 10^{6}Â units of volume. 10 ppm of SO_{2}Â in the air means 10 mL of SO_{2}Â is present in 10^{6}Â mL of air.

**9) Formality**

It is the number of mass in grams present per litre of solution. In case, formula mass is equal to molecular mass, formality is equal to molarity. Like molarity and normality, the formality is also dependent on temperature. It is used for ionic compounds in which there is no existence of a molecule. A mole of ionic compounds is called formal and molarity as the formality.

Where,

- w = weight of solute,
- f = formula weight of solute
- V= volume of solution
- n
_{f}= no. of gram formula weight

## A Solved Example For You

Q:Â What is the weight percentage of urea solution in which 10 gm of urea is dissolved in 90 gm water.

Solution:**Â **Weight percentage of urea = (weight of urea/ weight of solution)Â 100

= 10/(90+10)Â Â 100 = 10% urea solution (w/W)

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