Solutions – Definition, Examples, Properties and Types

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What is a Solution? 

A solution is a homogeneous mixture of two or more than two substances whose particle size is 0.1nm -1nm. Homogeneous means that the components of the mixture form a single phase. 

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You must have seen many types of solutions in general such as soda water, sharbat, salt solution etc. You would have seen brass utensils as well which are also homogeneous solutions of solid into solid. We can prepare solutions of solid-liquid, solid-solid, solid-gas, liquid-solid, liquid-liquid, gas-solid, gas-liquid and gas-gas. Gasoline, alloys, air, starch solution etc. are examples of solutions. 

Solutions consist of -Solute and Solvent. Let’s discuss solute and solvent in detail. 

Solute – The substance which is being dissolved in solvent to form a solution is called a solute. It is present in lesser quantity than solvent in the solution.

Solvent – The component of a solution in which solute is being dissolved is called a solvent. It is present in more quantity than solute in solution. 

For Example - If we take a solution of salt-water. Then in this solution salt is solute and water is a solvent as salt is being dissolved in water and salt is present in lesser quantity than water in the solution. 

Thus, we can say solvent forms a major proportion of the solution. 


Examples of Solutions 

Sugar-water, salt solution, brass, alloys, alcohol in water, aerosol, air, aerated drinks such as Coca-Cola etc. are the examples of solutions. When we work with chemistry, we generally prepare many types of solutions such as copper in water, iodine in alcohol etc. 


Properties of a Solution 

A solution possesses following properties – 

  • A solution is a homogeneous mixture. 

  • The constituent particles of a solution are smaller than 10-9 metre in diameter. 

  • Constituent particles of a solution cannot be seen by naked eyes. 

  • Solutions do not scatter a beam of light passing through it. So, the path of the light beam is not visible in solutions. 

  • Solute particles cannot be separated by filtration. 

  • Solute or solvent particles do not settle down when left undisturbed. 

  • Solutions are stable at a given temperature. 


Types of Solutions 

Solutions can be divided into following types of the basis of quantity of solute in a solution – 

Unsaturated Solution – The solution in which we can add more solute at a given temperature is called unsaturated solution. 

Saturated Solution – The solution in which we cannot dissolve more solute in the solvent at a given temperature is called saturated solution. 

Supersaturated Solution – The saturated solution in which we add more solute by raising the temperature or pressure is called a supersaturated solution. In these solutions generally crystals start forming. 

Solutions can be divided into following types of the basis of amount of solvent added- 

Concentrated Solution – The solution in which solute is present in large quantity is called a concentrated solution. 

Dilute Solution – it has a very small amount of solute in a large quantity of solvent. 


Expressing Concentrations of Solutions 

Concentration of a solution can be expressed quantitatively and qualitatively. Qualitatively it can be expressed as a dilute solution or concentration solution. Quantitatively it can be expressed by mass percentage, volume percentage, parts per million etc. 


Mass Percentage (w/w) -

Mass Percentage of a Component = \[\frac{Mass \; of \; the \; Component \; in \; the \; Solution}{ Total \; Mass \; of \; the \; Solution}\] x 100


Volume Percentage (V/V) –

Volume Percentage of a Component = \[\frac{Volume \; of \; the \; Component}{ Total \; Volume\; of \; the \; Solution}\] x 100


Mass by Volume Percentage (w/V) – Mass of Solute Dissolved in 100mL of the Solution.

Parts per Million (ppm) – 

Parts Per Million = \[\frac{Number \; of \; Parts \; of \; the \; Component}{Total \; Number \;of \; Parts \; of \; all \; Components \; of \; the \; Solution}\] x 106


Mole Fraction (x) –

Mole Fraction of a Component = \[\frac{Number \; of \; Moles\; of \; the \; Component}{Total \; Number \;of \; Moles\; of \; all \; Components \; of \; the \; Solution}\] 


Molarity (M) – Number of Moles of Solute Dissolved in One Liter of Solution. 

Molarity =  \[\frac{Moles \; of \; Solute}{Volume \; of \; Solution \; in \; Litre}\]


Molality (m) – Number of Moles of Solute in One KG of the Solvent. 

Molality =  \[\frac{Moles \; of \; Solute}{Mass \; of \; Solvent \; in \; Kg}\]


Solubility 

Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specific temperature. 

Following Factors Affect the Solubility of a Substance –

  • Nature of solute 

  • Nature of solvent 

  • Temperature 

  • Pressure 

Solubility of a Solid in a Liquid – Polar solutes get dissolved in the polar solvents while non – polar solutes get dissolved in the non – polar solvents. It means all solids do not get dissolved in all types of liquids. Polar solid compounds will dissolve in polar liquid solvents while non – polar solid compounds will dissolve in non – polar liquid solvents.  

Solubility of a solid into a suitable liquid is affected by temperature changes. Dynamic equilibrium in these type solutions must follow Le Chateliers principle. 

  • For a nearly saturated solution, the dissolution process is endothermic (∆H > 0). In this condition the solubility increases with increase in temperature. 

  • For the exothermic process (∆H < 0), the solubility decreases with increase in temperature. 

Solids and liquids are incompressible or very lightly compressible, so they are almost remaining unaffected by change in pressure. That’s why solubility of a solid in a liquid remains unaffected by pressure. 

Solubility of a Gas in a Liquid – Solubility of a gas in a liquid gets greatly affected by the pressure and temperature. 

Henry’s Law – The Henry’s law was given by English Chemist William Henry who first formulated the quantitative relation between pressure and solubility of a gas in a solvent. 

The law states that ‘at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.’

Dalton proved that the solubility of a gas in a liquid solution is a function of partial pressure of the gas. 

Mole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution. 

Another form of Henry’s law states that ‘the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.’ It can also be expressed as –

p = KH x

where p = partial pressure of the gas 

KH = Henry’s constant 

x = mole fraction of the gas 

At a given pressure if the value of KH increases for various gases then their solubility decreases accordingly. 


Applications of Henry’s law – It is applied in the production of carbonated beverages to increase the solubility of carbon dioxide gas in beverages. 

People who live in high altitudes have low concentration of oxygen in their blood and feel weak. This condition is called hypoxia. It is observed in the climbers as well who climb up to high altitudes. 

When scuba divers go underwater, they breathe air at high pressure. Increased pressure increases the solubility of gases in the blood. High concentrations of nitrogen and other gases can be lethal. 

Dissolution is an exothermic process, so the solubility decreases with increase in temperature. 


Raoult’s Law – It was given by French Chemist Francois Marte Raoult in 1886. Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

If we take a binary solution of two volatile liquids and those two components are denoted by component A and B. Then, for component A –

PA ∝ xA

PA = p xA

Where, PA = partial vapour pressure of the component of A

xA = mole fraction of component A 

pA° = vapor pressure of the pure component A at the same temperature. 

For component B –

PB = p xB

Where, PB = partial vapour pressure of the component of B

xB = mole fraction of component B

PB° = vapor pressure of the pure component B at the same temperature. 

According to Dalton’s law of partial pressures, the total pressure (ptotal) over the solution phase in the container will be the sum of the partial pressures of the components of the solution. So, it can be written as –

Ptotal  = pA + pB

On substituting the values of pA and pB

Ptotal = p xA + p xB

Ptotal = (1 - xB) p + p xB

Ptotal = p + (p- p)xB


Following Conclusions can be Drawn from the Above Equation 

  1. Total vapour pressure over the solution can be related to the mole fraction of any one component of the solution. 

  2. Total vapour pressure over the solution varies linearly with the mole fraction of component B. 

  3. Depending on the vapour pressures of the pure components A and B, total vapour pressure over the solution decreases or increases with the increase of the mole fraction of component A.


Raoult’s Law can Also be Stated as Follows 

The partial pressure of each volatile component (or gas) in the solution is directly proportional to its mole fraction. 

On the basis of Raoult’s law, liquid – liquid solution can be classified into following two types –

  • Ideal solutions

  • Non – ideal solutions 

Ideal solutions – Those solutions which obey the Raoult’s law over the entire range of concentration are known as ideal solutions. For ideal solutions – 

mixH = 0

mixV = 0

Non – Ideal Solution - Those solutions which do not obey the Raoult’s law over the entire range of concentration are known as non - ideal solutions. The vapour pressure of the non - ideal solution is either higher or lower than that predicted by Raoult's law. If it's higher than that predicted by Raoult’s law then, the solution exhibits positive deviation. If it's lower, then the solution exhibits negative deviation from the Raoult’s law. 

Solution shows positive deviation due to weaker intermolecular interactions or forces between solute – solvent molecules than solute – solute and solvent – solvent molecules. This weaker interaction increases the vapor pressure and results in positive deviation than Raoult’s Law. 

Solution shows negative deviation due to weaker intermolecular interactions or forces between solute – solute and solvent – solvent than solute –solvent molecules. This weaker interaction decreases the vapor pressure and results in positive deviation than Raoult’s Law. 

  • When only solvent molecules are volatile and present in vapor phase (solute molecules are non - volatile) then only solvent molecules contribute to vapor pressure. If p1 is the vapor pressure of the solvent, x1 is its mole fraction and p1° is its vapor pressure in the pure state, then according to Raoult’s law –

P1 ∝ x1

As the proportionality constant is equal to the vapor pressure of solvent at its pure state. So, we can write –

P1 = x1 p


Colligative Properties – Those properties which depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution, are called colligative properties. 

Relative Lowering of Vapor Pressure – Raoult’s established that lowering of vapor pressure does not depend on the identity of solute particles instead it only depends on the concentration of solute particles. We can write –

P1 = x1 p

Vapour pressure of the pure solvent will be more than that of the solvent. So, we can write change in vapor pressure as follows –

∆p1 = p  - p1 

= p- p  x1

= p(1 - x1)

As we know x2 = 1 - x1 so, we can write –

∆p1 = x2 p

If a solution contains many non – volatile solutes, then the lowering of the vapor pressure depends on the sum of mole fraction of different solutes. So, the above equation can be written as –

∆p1/p1° = (p1°- p1)/p1° = x2 ---------(1)

In the above equation the left - hand side equation is called relative lowering of vapor pressure which is equal to the mole fraction of the solute. 

As x2 = n2\(n1+n2)so, the equation (1) can be written as follows –

(p1°- p1)/p1° = n2\(n1+n2)

n1 = number of moles of solvent in the solution 

n2 = number of moles of solute in the solution 

For a highly diluted solution, n1 > > n2, so n2 can be neglected as it’s a very small value. Thus, we can write –

(p1°- p1)/p1° = n2/n1

As we know number of moles = mass/molar mass so, we can write –

(p1°- p1)/p1° = (w2/M2)/(w1/M1)

(p1°- p1)/p1° = (w2/M2) x (M1/w1)

Where, w1 and M1 are the mass and molar mass of solvent while w2 and M2 are the mass and molar mass of solute. 


Elevation of Boiling Point – Elevation of boiling point also depends only on the number of solute particles instead of nature of the solute particles. 

If boiling point of the pure solvent is Tb° and the boiling point of solution is Tb. Elevation of boiling point will be –

Tb = Tb - Tb° 

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According to the results of the experiments, for dilute solutions, elevation of the boiling point is directly proportional to the molal concentration of the solute in a solution.  Thus, we can write –

∆T∝ m 

On removing the proportionality –

∆Tb = Kb m

Where, m = molality 

Kb = Boiling point elevation constant or molal elevation constant or Ebullioscopic constant. Its unit is K kg/ mol.

If w2 and M2 are the mass and molar mass of solute which are dissolved in the w1 gram of solvent. 

m = (w2/M2)/(w1/1000) = (1000 x w2)/(M2 x w1)

∆Tb = (Kb x 1000 x w2)/(M2 x w1)

M2 = (Kb x 1000 x w2)/(∆Tbx w1)


Depression of Freezing Point – Solution shows depression of freezing point compared to pure solvent. 

Freezing Point – Freezing point can be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.

When we add some non – volatile solid in the solvent, its vapor pressure decreases (Raoult’s law). Due to decrease in vapor pressure, it becomes equal to solid at lower temperature. Therefore, the freezing point of the solvent decreases.  

Depression in freezing point can expressed as -

∆Tf= Tf* - Tf 

Where, Tf* = freezing point of pure solvent 

Tf = freezing point of the solvent when non - volatile solute is dissolved in it.

  • Depression of freezing point for dilute solution is directly proportional to molality of the solution. It can be expressed as –

∆Tf ∝ m  

∆Tf = Kf m

Where, ∆Tf = depression of freezing point 

m = molality 

Kf = Freezing point depression constant or molal depression constant or cryoscopic constant

If w2 and M2 are the mass and molar mass of solute which are dissolved in the w1 gram of solvent. 

m = (w2/M2)/(w1/1000) = (1000 x w2)/(M2 x w1)

∆Tf = (Kf x 1000 x w2)/(M2 x w1)

M2 = (Kf x 1000 x w2)/(∆Tfx w1)


Osmosis and Osmotic Pressure – The process of movement of the solvent across a semipermeable membrane towards a higher concentration of solute is called osmosis. Osmotic pressure is the minimum pressure required or applied to a solution to halt the flow of its pure solvent across a semipermeable membrane. Osmotic pressure is also a colligative property. It depends on the concentration of the solute in the solution. It is expressed as follows – 

π = iCRT 

Where, π = osmotic pressure 

i = van’t Hoff factor

C = molar concentration of the solute in the solution 

R = universal gas constant 

T = temperature 

When the pressure is applied more than that of osmotic pressure, then pure solvent starts flowing out of the solution through the semipermeable membrane. This phenomenon is called reverse osmosis. 

Experimental values of molar masses sometimes differ from the theoretical values of molecular masses (calculated from the colligative properties of solutions). These values are known as abnormal molar masses. 

This ends our coverage on the summary of the unit “Solutions”. We hope you enjoyed learning and were able to grasp the concepts. You can get separate articles as well on various subtopics of this unit such as characteristics of Types of solutions, Raoult’s Law etc. on Vedantu website. We hope after reading this article you will be able to solve problems based on the topic. We have already provided detailed study notes or revision notes for this unit, which you can easily download by registering yourself on Vedantu website. Here in this article we have discussed the unit in a summarized way with the emphasis on important topics of the unit.  If you are looking for solutions of NCERT Textbook problems based on this topic, then log on to Vedantu website or download Vedantu Learning App. By doing so, you will be able to access free PDFs of NCERT Solutions as well as Revision notes, Mock Tests and much more.