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Concentration of a Solution

A solution is a homogenous mixture of two or more pure substances whose composition may be altered within certain limits. Though the solution is homogenous in nature, yet it retains the properties of its constituents. Generally, solution is composed of two components, solute and solvent. Such type of solution is known as binary solutions. Solvent is that component in solution whose physical state is the same as that of the resulting solution while other component is called a solute. If the physical state of both components is same than the component in excess is known as solvent and other one is called a solute. Each component in a binary solution can be in any physical state such as liquid, solid and gaseous state. Some common type of solutions is as follows.  

Types of Solution 
Solute
Solvent  Examples
 Gaseous Solution   Solid  Gas   Sub limitation of Solid in Gas; Camphor Vapour in $N_2$ gas.
Water Vapours in air(mist),Mixture of gas ,air.
 Liquid
 Gas
 Liquid Solution Solid Liquid Salt in Water, Sucrose in Water,alcohol in Water,$CO_2$ in water
Aerated drink. 
 Liquid
 Gas
 Solid Solution Solid Solid Alloys, Adsorption of Gas on Metal ,Solution of Hydrogen in  Palladium.
 Liquid
 Gas

Related Calculators
concentration dilution calculator Molar Concentration Calculator
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Finding the Concentration of a Solution

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The concentration of a solution is the amount of solute dissolved in a known amount of the solvent or solution. Solution can be described as dilute or concentrated solution as per their concentration. A dilute solution has a very small quantity of solute while concentrate solution has a large quantity of solute in solution. Various concentration terms are as follows.

1. Mass percentage


It may be defined as the number of parts of mass of solute per hundred parts by mass of solution. If WB is the mass of solute B and Wa is the mass of solvent A, then
Mass % = $\frac{W_B}{W_A + W_B}$ x 100
For example, the percent composition by mass of a 100 g sugar solution which contains 15 g sugar will be

$\frac{15g\ Sugar}{100g\ solution}$
x 100 = 15% Sugar solution


2. Volume Percent


It can be represented as % v/v or % volume and used to prepare such solutions in which both components are in liquids state. It is the number of parts of by volume of solute per hundred parts by volume of solution. Therefore,
Volume % = $\frac{volume\ of\ solute}{total\ volume\ of\ solution}$ x 100%
For example, 20% solution of ethanol by volume means that 100 cm3 of the solution contains 20 cm3 of ethanol and 80cm3 of water.

3. Mass-volume percentage (W/V %)


It may be defined as the mass of solute present in 100cm3 of solution. For example, If 100 cm3 of solution contains 5g of sodium hydroxide, than the mass-volume percentage will be 5% solution.

4. Parts per million (ppm)


The very low concentration of solute in solution can be expressed in ppm. It is the numbers of parts by mass of solute per million parts by mass of the solution.
Parts per million (ppm) = $\frac{Mass\ of\ solute X 10^6}{Mass\ of\ solution}$

5. Mole Fraction (X)


Mole fraction may be defined as the ratio of number of moles of one component to the total number of moles of all the components (solute and solvent) present in solution. It is denoted by letter x and the sum of all mole fractions in a solution always equals one. If a solution contains the components A and B and WA g of A and WB g of B are present in it, than the moles of A (nA) = WA/(MA);
Moles of B(nB) = WB/(MB)
Where MA and MB are molar mass of A and B respectively 
Total number of moles of A and B = nA + nB
Hence Mole fraction of A (xA) = nA / nA + nB 
Mole fraction of B (xB) = nB / nA + nB 
xA + xB = ( nA / nA + nB ) +( nB / nA + nB )= 1
Mole fraction does not depends upon temperature and can be extended to solutions having more than two components. 

6. Molarity (M)


Molarity is most common unit for concentration of solution. It is defined as the number of gram mole of solute present in one litre or one dm3 of the solution or millimol of solute present in one mL of solution. 
Mathematically, it can be expressed as
M = $\frac{Mole\ of\ solute}{Volume\ of\ solution(L)}$
  • Since mole of solute = Mass of solute in gram/ Molar mass of solute
  • Therefore, M = Mass of solute in gram/ Molar mass of solute x Volume of solution (L)
  • If nB moles of solute B are present in V liter of solution.
Molarity = $\frac{n_B}{V(L)}$
Or
M = $\frac{W_B X 1000}{M_B X V(ml)}$
Where MB is the molar mass of solute and WB is the gram of solute present in V(ml) of solution. One molar solution is defined as the solution having unity Molarity. Such solutions have one mole of solute per liter of solution. The unit of Molarity is mol/L or mol dm3-

7. Molality(m)


The number of gram mole of the solute present in 1000 g of the solvent is known as molality of solution. It represented by letter ‘m’.

Molality (m) = $\frac{Mole\ of\ solute}{Mass\ of\ solvent(kg)}$
Or
m = $\frac{Mass\ of\ solute(g) X 1000}{Molar\ mass\ of\ solute X mass\ of\ solvent(g)}$
The unit of molality is mol/kg and it does not effect by temperature.

8. Normality(N)


The number of gram-equivalent of the solute present in one liter of solution is known as normality of solution. In other words; number of milli-equivalents of solute present in one mL of solution is called as normality of that solution. It represented by letter ‘N’.
Normality (N) = $\frac{gram\ equivalent\ of\ the\ solute}{volume\ of\ solution(L)}$
A solution having normality equal to unity is known as a normal solution. Such solution contains one gram equivalent of solute per dm3 of solution.

Calculating the Concentration of a Solution

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Concentration of solution can be calculated in terms of normality, Molarity, molality or mole fraction.

Formula for Concentration of a Solution

Solved Examples

Question 1: Calculate the mole fractions of the components of the solution composed by 92 g glycerol and 90 g water?
(M(water) = 18; M(glycerol) = 92)
Solution:
 
Moles of water = 90 g x 1 mol / 18 g = 5 mol water
Moles of glycerol = 92 g x 1 mol / 92 g = 1 mol glycerol
Total moles in solution = 5 + 1 = 6 mol
Mole fraction of water xwater = 5 mol / 6 mol = 0.833
Mole fraction of glycerol xglycerol = 1 mol / 6 mol = 0.167
 

Question 2: What will be the Molarity of solution when water is added to 10 g CaCO3 to make 100 mL of solution?
Solution:
 
10 g CaCO3 / (100 g CaCO3 / mol CaCO3) = 0.10 mol CaCO3
100 mL x 1 L / 1000 mL = 0.10 L
Molarity = Mole of solute / Volume of solution (L)= 0.10 mol / 0.10 L
Therefore; Molarity of given solution = 1.0 M
 

Question 3: Calculate the molality of a solution containing 20 g of sodium hydroxide (NaOH) in 250 g of water?
Solution:
 
Moles of sodium hydroxide= 20 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.2 mol NaOH
250 gm = 0.25 kg of water (250 g water x 1 kg / 1000 g = 0.25 kg)

Hence molality of solution = Mole of solute/ Mass of solvent (kg)= 0.2 mol / 0.25 kg
or Molality(m) = 0.8M / kg or 0.8 m

 

Question 4: Calculate the number of grams of copper sulphate (CuSO4) needed to prepare 250.0 mL of 1.00 M CuSO4?
Solution:
 
Molarity = $\frac{Mass\ of\ copper\ sulphate\ (gm) X 1000}{Molar\ mass\ of\ copper\ sulphate\ X\ V (ml)}$
                                    
Molar mass of copper sulphate = 159.6 g/mol
V =250 ml
Molarity =2.00M
 
Hence Mass of copper sulphate (gm) =  $\frac{Molarity X Molar\ mass\ of\ copper\ sulphate\ X\ V (ml)}{1000}$
                                                                                         
= $\frac{2.00\ X\ 159.6\ X\ 250}{1000}$
                                                                 
= 79.8 gm of Copper sulphate

 

Question 5: Calculate the normality of solution containing 296.4 grams of Ca(OH)2 in 600 ml of water.
Solution:
 
Normality = $\frac{mass\ of\ solute (gm) X 1000}{Equivalent\ mass\ of\ solute X V (ml)}$
                
Equivalent mass of calcium hydroxide = Molar mass/ acidity
Molar mass of calcium hydroxide= 74.1 amu
Acidity =2 (as Ca(OH) contains 2 replaceable hydroxides)
Therefore Equivalent mass of calcium hydroxide = 74.1/2=37.05
 
Normality of solution = $\frac{296.4 X 1000}{37.5 X 600}$
= 13.3 N

                                         
 

Question 6: Calculate the ppm of mercury in water in given sample contain 30 mg of Hg in 500 ml of solution.
Solution:
 
Parts per million = $\frac{Mass\ of\ solute X 10^6}{Mass\ of\ solution}$
                                                     
Mass of Hg = 30 mg
Mass of water = 500/1 = 500g = 50 x 104 mg
(As Mass = Volume /density; density of water 1 g/ml)
 
Therefore

ppm of mercury =
$\frac{30mg\ of\ lead X 10^6}{50 X 10^4 of\ solution}$
= 60 ppm of mercury

 


Concentration of Ions in a Solution

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In general, we are using ionic compounds in various reactions. Ionic compounds are good conductor in their aqueous solution due to the presence of ions in solution. Therefore in ionic compounds the determination of concentration of ions is more useful than concentration of solution. For determination of concentration of ions in solution, first we have to write balance chemical equation that show how these substances break down into ions. 

For example, the dissociation of potassium carbonate in aqueous can be written as follows.

K2CO3 (s) $\to$ 2 K+(aq)+ CO32-(aq)

Every one mole of K2CO3 produces two moles of K+ ions and two moles of carbonate ions in solution with no change in volume. Now if we have 0.40 M solution of potassium carbonate ( K2CO3 ), the concentration of both ions. K+ and CO32- can be calculated by using the coefficients in our balanced equation.

Concentration of potassium ion, [K+] = 2 × [ K2CO3 ] = 2 × 0.40 M = 0.80 M
Concentration of carbonate ion [ CO32- ] = 1 × [ K2CO3 ] = 1 × 0.40 M = 0.40 M

Let’s calculate the concentration of ions in 0.30 M solution of Iron (III) nitrate. The balanced chemical equation of dissociation of Iron (III) nitrate in its aqueous solution is as follow;

Fe(NO3)3 $\to$ Fe3+(aq) + 3 NO3-(aq)

The coefficients from the balanced chemical equation may be used to determine the concentration of the ions.

Concentration of [ Fe3+ ] = 1× [ Fe(NO3)3 ] = 1 × 0.30= 0.30 M
Concentration of [ NO3- ] = 3× [ Fe(NO3)3 ] = 3 × 0.30 = 0.90 M
 Example 1

Calculate ion concentrations in a 3.00 L solution containing 17.1 g aluminium sulphate, Al2(SO4)3. (Atomic mass of Al =27.0 g/mol, O=16 g/mol, S =32.0 g/mol)

Calculation :

Molar mass of Aluminium sulphate = 2 × 27.0 +3× 32.0.0+12 × 16.0 = 342.0 g/mol
Concentration of Al2(SO4)3 = Mass of Al2(SO4)3 / Molar mass of Al2(SO4)3 x V(L)

= 17.1 /342.0 x 3 = 0.0166 M
Dissociation of aluminium sulphate in its aqueous solution can be written as

Al2(SO4)3 $\rightarrow$ 2Al3+ (aq) + 3SO42- (aq)

From balanced equation of dissociation;

Concentration of [Al3+] = 2× [ Al2(SO4)3 ] = 2 × 0.0166 = 0.0333 M or 3.33 ×10-2M
Concentration of [SO42-] = 3× [ Al2(SO4)3 ] = 3 × 0.0166 = 0.0498 M or 4.98×10-2M
 Example 2

What is the concentration of sodium and sulphate ions in a 0.650 M solution of Na2SO4?

Calculation:

The dissociation of sodium sulphate in its aqueous solution can be written as;

Na2SO4(s) $\rightarrow$ 2Na+(aq) + SO42-(aq)

The concentration of sodium and sulphate ions will be

Concentration of Sodium ions; [Na+] = 0.650 M Na2SO4 x 2Na+ /1 Na2SO4 = 1.3 M Na+
Concentration of sulphate ion; [SO42-] = 0.650 M Na2SO4 x1SO42- /1 Na2SO4 = 0.650 M SO42-

Like Molarity is used to express the concentration of solution, formality is used to calculate the concentration of ions in solution. It is the number of formula weight units of solute present in one liter of solution. Formality is represented by letter ‘F’. The mass of one mole of a compound equals to the formula weight in grams. Formality is used for the determination of the number of moles of a compound from the number of moles of ions present in solutions of ionic compounds.

F = $\frac{n Formula\ weight\ units}{ Volume\ of\ solution\ (L)}$

Determining the Concentration of a Solution Beer's Law

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The properties of light can be described by wavelength and frequency. On the basis of wavelength and frequency, light can be classified into various types. Out of these radiations, only visible light can be seen by human eyes. The wavelength of visible light lies between 400 nm to 750 nm. 

The visible spectrum is composed of seven colors, where each color associated with certain wavelength. When visible light falls on the surface of any substance, it absorbs and complimentary color transmitted. The color of object is due to complementary color which is just opposite to absorbed color. 

The mathematical relation between absorption of light with concentration of solution may be given by Beer-Lambert Law.
A = ebc
Where ,
A = Absorbance of light at given wavelength<
e = Molar absorptivity(L mol-1 cm-1)
b = Path length through the solution that the light has to travel (cm)
c = Concentration of the solution (mol/liter)

As the absorption of a particular color of light increases, the color of solution increases in that region of spectrum which is not absorbed by solution. The absorbance (A) is directly proportional to the concentration of solution, therefore increases with increasing the concentration of solution. 

For the determination of concentration of solution of an unknown sample, the absorption of a series of known solution is measured and plotted which is called as Beer’s law plot or calibration curve. This curve can be used for the determination of concentration of unknown sample by measuring the absorbance at same wavelength and path length.


Beers Law
Colorimetry is used for the determination of concentration of solution which based on the relation between absorption and path length. Since molar absorptivity is unique for each absorbing species and vary with wavelength, therefore any sum of path length and concentration will be same when multiplied will absorb the same amount of light. Absorbance is related to appearance of solution, therefore two solutions (A1 and A2) with same color will show same absorbance.

A1 = A2

From Beer-Lambert Law

A = e bc,
e 1 = e 2 (for same substance)
Therefore
b1c1 = b2c2
Or
b1c1 = b2c2 ........(2)

The concentration of unknown solution can be determined by using equation (2).

Where
  • A1= absorbance of known solution
  • e 1= Molar absorptivity of known solution
  • c1= concentration of known solution
  • A2= absorbance of unknown solution
  • e 2= Molar absorptivity of unknown solution
  • C2= concentration of unknown solution

colorimeter

For example concentration of unknown solution of nickel sulphate can be determined by using a known solution of same compound in colorimeter. This solution is known as standard solution and must have deep green color. A colorimeter consists of a red light from LED source which passes through solution and strike to photocell. 

Since slandered solution of nickel sulphate is deep in color, therefore more concentrated and absorbed more light than a solution of low concentration. This transmitted light is monitored by photocell. Measure the absorbance of a series of known solution prepared by using standard solution and plots a graph between absorbance and concentration of solutions. The graph shows a straight line with positive slop as according to Beer-Lambert law.

Similarly measure the absorbance of unknown nickel sulphate solution with the colorimeter. Now we can locate the absorbance of this solution on the vertical axis of the graph and the corresponding concentration can be found on the horizontal axis.
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