In the field of Chemistry, the term concentration is the amount of solute present in the Solution. Concentration is a very important concept used to describe the quantity of solute dissolved in the solution. Various Concentration terms exist by which the concentration measure which includes the molality, molarity, and percentage composition. This concentration concept was discovered and evolved with time over the years. Chemists like Antoine Lavoisier and Joseph Proust initially laid the foundation for concentration to understand solutions and chemical reactions.
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Concentration also helps in controlling the concentration of reactant for the desired product by ensuring safety in laboratory and industrial settings. Correct concentration measurements are also important for the stoichiometric calculation to determine the number of reactants and products in chemical reactions and also for preparing the standard solution as well
The concentration of a solution gives us an idea about the relative amount of solute and solvent present in the solution. The concentration can be expressed either qualitatively or quantitatively. For example, qualitatively we can say that the solution is dilute (i.e., a relatively very small quantity of solute) or it is concentrated (i.e., a relatively very large quantity of solute). But in reality, the qualitative description can cause confusion, and hence there is a need for a quantitative description of the solution.
There are several ways by which we can describe the concentration of the solution quantitatively.
(1) Mass percentage (w/w):
It is the mass of any component present in 100 g of solution.
Mathematically, it can be defined as:
Mass % of a component = Mass of the component in the solution Total mass of the solution ×100
For example, a solution described as 20% by mass of glucose in water, means that 20 g of glucose is dissolved in 80 g of water resulting in a 100 g solution.
The mass % can also be expressed in terms of the mass fraction by simply removing the 100 from the above-given formula
Concentration described by mass percentage is commonly used in industrial chemical applications.
(2) Volume percentage (V/V):
It is the volume of any solute present in 100 ml of the solution. Mathematically it is defined as:
Volume % of a component = Volume of the component Total volume of solution ×100
For example, a 20% Methanol solution in water means that 20 mL of Methanol is dissolved in water such that the total volume of the solution is 100 mL. Solutions containing liquids are commonly expressed in this unit.
(3) Mass by volume percentage (w/V):
It is the mass of solute dissolved in 100 mL of the solution. Mathematically, it is defined as:
Mass by Volume % of a component = Mass of the component Total volume of solution ×100
For example, a 20% weight-by-volume solution of Glucose in water means that 20 g of Glucose was dissolved in water to obtain a 100ml solution.
This concentration term is commonly used in medicine and pharmacy.
(4) Parts per million (ppm):
When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm) and is defined as: Parts per million = Number of parts of the component Total number of parts of all components of the solution ×106
As in the case of percentage, concentration in parts per million can also be expressed as mass to mass, volume to volume, and mass to volume.
This is generally used in expressing the hardness of water and in expressing the concentration of dissolved oxygen in water etc.
For example, if the hardness of a hard water sample is 100pm in CaCO3, it means that 100 g of CaCO3 is present in 106 g of the water sample.
(5) Mole fraction:
It is the ratio of the moles of any component present in the solution to the total moles present in solution. A commonly used symbol for mole fraction is X and the subscript used on the right-hand side of X denotes the component.
It is defined as: Mole fraction of a component = Number of moles of the component Total number of moles of all the components
For example, in a binary mixture, if the number of moles of A and B is nA and nB respectively, the mole fraction of A will be:
xi=n1n1+n2+……+ni=ni∑ni
It can be shown that in a given solution sum of all the mole fractions is unity, i.e.
x1+x2+……………+xi=1
The mole fraction unit is very useful in relating some physical properties of solutions, say vapor pressure with the concentration of the solution, and quite useful in describing the calculations involving gas mixtures.
(6) Molality(m):
It is defined as the number of moles of the solute present per kilogram (kg) of the solvent and is expressed as:
Molality (m)= Moles of solute Mass of solvent in kg
For example, 1 molal solution of NaOH means that 1 mol (40 g) of NaOH is dissolved in 1 kg of water.
(7) Molarity (M):
It is defined as the number of moles of solute dissolved in one liter of solution
Molarity = Moles of solute Volume of solution in litre
For example, 0.5 mol L-1 (or 0.5 M) solution of NaOH means that there is 0.5 mol of NaOH dissolved in water to obtain one liter of solution.
Each method of expressing the concentration of the solutions has its own merits and demerits. Mass %, ppm, mole fraction, and molality are independent of temperature, whereas molarity is a function of temperature. This is because volume depends on temperature and mass does not.
Example.1
1. 25 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35 ml. The molarity of barium hydroxide solution was
1) (correct)0.07
2)0.14
3)0.28
4)0.35
Solution
Let M1 = molarity and V1 =Volume of barium hydroxide solution.
M2 = molarity and V2 = Volume of hydrochloric acid solution.
Now, we knowM1V1n1=M2V2n◻
n1 = 2 (n factor of Ba(OH)2) and n2 = 1 (n factor of HCl)
So, so, 2×M1 V1=M2 V22×M1×25=0.1×35M1=0.1×3525×2=0.07
Hence, the answer is the option (1).
Example.2
2. The molarity of 0.006 mole of NaCl in 100ml solution is
1) (correct)0.06
2)0.6
3)0.006
4)0.066
Solution
As we learned
Molarity -
Molarity = Moles of solute Vol.of solution (L)
M=nV(l)=0.0060.1=0.06
Hence, the answer is (0.06M).
Example.3
3.9.8 g of H2SO4 is present in 2 litres of a solution. The molarity (in M) of the solution is
1) (correct)0.05
2)0.1
3)0.2
4)0.01
Solution
Molality = Moles of solute Mass of solution (Kg) Molarity =% by mass ×10×d Gram Molecular Mass Molarity =22×10×1.253342=0.805M Normality =% by mass ×10×d Gram Equivalent Mass Normality =22×10×1.253342/6=4.83 N Molality =22×1000342(100−22)=0.825 m
Hence, the answer is the option (1).
Example.4
4. The mole fraction of the solute in one molal aqueous solution is
1) (correct)0.018
2)0.036
3)0.027
4)0.009
Solution
As we learned
Mole Fraction -
Mole Fraction = Moles of solute Moles of solute + Moles of solvent W=1000gm(H2O);n=1 moleN=WM=100018=55.55Xsolute =nn+N=11+55.55=0.018.
Hence, the answer is (0.018).
Example.5
5.A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made up 100 mL . The mass of calcium sulphate formed and the concentration of OH−in resulting solution, respectively , ar (Molar mass of Ca(OH)2,Na2SO4 and CaSO4 are 7 143 and 136 g mol−1, respectively ; Ksp of Ca(OH)2 is 5.5×10−6 )
1) (correct) 1.9 g,0.28 mol L−1
2) 13.6 g,0.28molL−1
3) 1.9 g,0.14molL−1
4) 13.6 g,0.14molL−1
Solution
Given,Mol of Na2SO4 = 2/142 = 14 m mol
Ca(OH)2+Na2SO4⟶CaSO4+2NaOHmmol1001414 m/mol28 m/mol
Mass of CaSO4=14×1361000=1.9gm
Molarity of OH−=28100=0.28 mol/L
Example.6
6.10.30mgO2 is dissolved into a liter of sea water of density 1.03 g/mL. The concentration of O2 in ppm is:
1) ( correct )10
2) 20
3) 25
4) 40
Solution
1030 g of sea water contains =10.3×10−3 g
106gm of sea water contains =(10.3×10−3)/1030×
ppm=10.3×10−31030×106
=10.
Hence, the answer is the option (1).
The concept of concentration in solutions is very important in the feild of chemistry for industrial purposes and many more. The correct measure of the concentration is the main key to controlling the chemical reaction to make sure that reactants are mixed in the accurate proportion to get the desired products. This quality requires the scientists and engineers to predict the rate of reaction, equilibrium positions, and product yield. The lab's concentration is crucial for making the standard Solution by knowing their exact concentration which is important for setting the instruments and making sure that to get the desired results. This is useful for quantitative analysis and measuring the correct composition of various materials.
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