EXPRESSION OF CONCENTRATION OF SOLUTION

Concentration Terms

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 n_A and n_B 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.

Recommende topic video(Expression of concentration of solution)

Some Solved Examples

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).

Summary

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.