Content Writer | Updated On - Oct 28, 2024
Solutions can be defined as homogeneous mixtures of two or more components wherein the particle size is smaller than 1 nm. It is commonly applied to the liquid state of matter.
- Solutions can also be formed by gases and solids as well.
- All the components present in the mixture generally appear as a single phase.
- The solution always has particle homogeneity, thus indicating that all particles are evenly distributed.
- Solute and solvent are two components of the solution.
- Aqueous and non-aqueous are divided on the basis of solvent.
- The particles are very small and can only be seen by a microscope.
- Some common examples include sugar and salt in water , soda water, etc.
- Air and alloy are the most common examples of solutions in gases and solid states.
What is a Solution?
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Solution can be defined as a homogenous mixture of two or more substances. The substances can either be in the same physical state or different physical states.
- The substances forming the solution are known as the components of the solution.
- It is a mixture of various substances.
- The type of mixture is decided based on the number of components present in the solution.
- If it is made of two components, then it is a binary solution.
- The particles are not visible by the naked eye.
Characteristics of Solution
Some of the important characteristics of solutions are as follows:
- Typically, solvents can be found in a major proportion as opposed to solute.
- The amount of solute is generally less in quantity than the solvent.
- Solute and solvent can be found in any state of matter i.e. solid, liquid or gas.
- Solutions found in the liquid state comprise a solid, liquid or gas further dissolved in a liquid solvent.
- Some examples of solid and gaseous solutions are alloys and air respectively.
Concentration of a Solution
The given amount of solute present in a solution is known as the concentration of a solution. The proportion of solute and solvent in the given solution is not even. Based on the solute proportion, a solution can be:
- Diluted
- Concentrated
- Saturated
Concentration of Solution = \(\frac{Amount\ of\ solute}{Amount\ of\ solution}\)
or, Concentration of Solution = \(\frac{Amount\ of\ solute}{Amount\ of\ solvent}\)
Methods of expressing the concentration of a solution:
The composition of a solution can be described by expressing the concentration of the solution. It can be expressed either qualitatively or quantitatively. Qualitatively we can say the solution is diluted, concentrated, saturated, unsaturated, etc. Solution can be described quantitatively by following formulas:
- Mass Percentage (w/w):
Mass % of a component = \(\frac {Mass\ of\ component\ in\ solution}{Total\ mass\ of\ solution} \times 100\)
- Volume Percentage (V/V):
Volume % of a component = \(\frac {Volume\ of\ component\ in\ solution}{Total\ mass\ of\ solution} \times 100\)
- Mass by Volume Percentage (w/V):
Mass % of a component = \(\frac{Mass\ of\ component}{100\ mL\ solution}\)
- Part upper million (ppm)
Part upper million (ppm) = \(\frac{Number\ of\ parts\ of\ component}{Total\ Number\ of\ parts\ all\ Components\ of\ Solution} \times 10^6\)
- Mole Fraction of Component
Mole Fraction of Component = \(\frac{{Moles\ of\ component}}{{Total\ number\ of\ moles\ of\ all\ the\ component}}\)
Molarity is given as= \(\begin{array}{l}\frac{Number\;of\;moles\;of\;element}{Volume\;of\;solution\;in\;litres}\end{array}\)
Molality is given as= \(\begin{array}{l} \frac {Number \ of \ moles \ of \ solute}{Mass \ of \ solvent \ in \ kgs}\end{array}\)
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Components of Solution
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A solution typically comprises of two components which are as follows:
Solute
A solute is a substance that is present in a smaller amount. It is the substance that is being dissolved in the solvent.
Characteristics of Solute
Some important characteristics of solute are as follows:
- Solute particles in a solution are not visible to the naked eye.
- A solution does not allow light beams to scatter.
- A solution is stable in nature.
Solvent
A solvent is a substance that is present in a large amount. It is also known as the dissolving medium.
Characteristics of Solvent
Some important characteristics of solvent are as follows:
- It is typically a liquid but can also be found in the state of a solid, gas, or supercritical fluid.
- The quantity of solute dissolved in a distinct solvent volume changes with temperature.
- It includes paint thinners, nail polish removers, dry cleaning, glue solvents, and perfumes.
- The most common solvent is water since it has the ability to dissolve almost all solutes.
Solution
What is a Mixture?
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Mixtures can be defined as substances which comprise two or more types of matter. Blood and air are examples of mixtures. Depending on the nature of components and their distribution, mixtures can be further classified as homogeneous and heterogeneous mixtures.
Homogeneous Mixture
A mixture whose components are uniformly distributed is called a homogeneous mixture.
Heterogenous Mixture
A mixture whose distribution is non-uniform is called a heterogeneous mixture.
Properties of Solution
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Some of the properties of solutions are:
- A solution is a homogeneous mixture.
- The particles are generally very small and have a diameter which is less than 1 nm.
- They are invisible to the naked eye.
- Particles do not scatter when a beam of light passes via it; thus, the path of the light becomes invisible.
- Solutes are generally inseparable from the mixture; thus, they do not sediment.
- A solution is stable.
- Its components cannot be separated by the method of filtration.
Types of Solutions
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The solutions can be classified into four categories which are as follows:
On the basis of water as a Solvent
On the basis of water as a solvent, solution is divided into two categories which are:
Aqueous Solution
A solution is said to be an aqueous solution when the only solvent dissolved in the solution is water.
Non-Aqueous Solution
A solution is said to be an non-aqueous solution when the solvent dissolved in the solution is other than water.
On the basis of amount of Solute
On the basis of amount of solute, solution is divided into three categories which are:
Saturated Solution: A solution is said to be saturated when it cannot dissolve any more amount of solute in it at any given temperature.
Unsaturated Solution: A solution is said to be unsaturated when it can dissolve any more amount of solute in it at any given temperature.
Supersaturated Solution: A solution is said to be supersaturated when it can dissolve maximum amount of solute that can be dissolved in a solvent at any given temperature.
On the basis of amount of Solvent
On the basis of amount of solvent, solution is divided into two categories which are:
Concentrated Solution
A solution is said to be concentrated when the amount of solute is larger than the amount of solvent. Mango juice and dark colour tea are some common example of concentrated solution.
Dilute Solution
A solution is said to be dilute when the amount of solute is lesser than the amount of solvent. Light colour tea are some common example of concentrated solution.
On the basis of concentration of solute in two Solutions
On the basis of concentration of solute in two Solutions, it is divided into three categories which are:
Isotonic Solution
A solution is said to be isotonic when it has higher concentration of solute dissolve in it.
Hypertonic Solution
A solution is said to be hypertonic when concentration of solute is exactly same in both the solutions.
Hypotonic Solution
A solution is said to be hypotonic when it has lesser concentration of solute dissolve in it.
Thus, based on the physical states of solute and solvent, solutions can be further classified as:
Types of Solution | Solute | Solvent | Examples |
---|---|---|---|
Solid – solid | solid | solid | Alloys, including Brass, bronze and more. |
Solid – liquid | solid | liquid | Sugar or salt solution in water. |
Solid – gas | solid | gas | Substance sublimation such as iodine or camphor in the air. |
Liquid – solid | liquid | solid | Mercury in amalgamated zinc |
Liquid – liquid | liquid | liquid | Benzene in toluene or alcohol in water. |
Liquid – gas | liquid | gas | Water vapour present in the air. |
Gas – solid | gas | solid | Absorbed Hydrogen in Palladium |
Gas – liquid | gas | liquid | Aerated drinks |
Gas – gas | gas | gas | Gas mixture |
What is Solubility?
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Solubility is the ability of a solute to dissolve in a solvent to form a solution. The maximum amount of a solute that can be dissolved in a certain amount of solvent at a certain temperature is termed as the solubility of a solute at that temperature.
The solubility of a solute depends on the following factors:
- The temperature of the solution
- Nature of the solvent
- Nature of the solute
- Pressure (in case of gases)
Types of Solubility
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
Unsaturated solution
A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
Saturated solution
A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an saturated solution.
Supersaturated solution
A solution which contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.
Solubility
Henry’s law
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Henry's Law states, “the partial pressure (p) of the gas in the vapour phase is proportional to the mole fraction (x) of the gas in the solution”. This law was formulated by William Henry in the early 19th century. Henry’s Law is for the liquid and gas solution. It can be expressed as
p ∝ x
- Thus,
p = KH . x
Here,
- KH is the Henry’s Law constant
- p is the partial pressure of the gas
- x is the mole fraction of the gas
- Different gases have different values of KH at the same temperature.
Raoult’s Law
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Raoult's law was formulated by French chemist François-Marie Raoult in 1887. It is applicable for a solution of two volatile liquids. According to Raoult’s law: Partial vapour pressure of every component in a solution is directly proportional to the mole fraction of that component.
Thus, Raoult’s Law formula is:
P = P0 X
Where,
- P0 = Vapour pressure of pure liquid component 1 at the same temperature
- Χ = Mole fraction of component in solution.
- P = Vapour pressure of component in solution.
Colligative Properties
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The properties of a solution which depends only upon the number of solute particles in the solution and not on the nature of the solute particles are known as colligative properties. The important colligative properties are as follows:
Relative Lowering in Vapour Pressure
Relative lowering in vapour pressure can be defined as the ratio between the lowering in vapour pressure and vapour pressure of pure solvent. If the vapour pressure of a pure solvent in a solution is P0 and the vapour pressure of the solution is PS
Then the relative lowering of vapour pressure is given as
\(\frac{P_o - P_s}{P_o}\)
Elevation of Boiling Point
The temperature at which the vapour pressure is equal to atmospheric pressure is known as the boiling point of the liquid. The vapour pressure of a solution decreases when we add a non-volatile liquid to a pure solvent.
- To make vapour pressure and atmospheric pressure equal, we have to increase the temperature of the solution.
- The difference in the boiling point of the solution and the boiling point of the pure solvent is known as the elevation in boiling point.
- Let T0b = boiling point of the pure solvent
- Tb = boiling point of the solution
- So, the elevation in boiling point will be
ΔTb = T0b – Tb
- Since, there is a relation between elevation in boiling point and molality of the solute ΔTb ∝ m
ΔTb = kb m
- where, kb = molal elevation constant
- By substituting the value of ‘m’
ΔTb = 1000. kb. m2/ M2 . m1
- Where,
- m2 = mass of solvent in g
- m2 = mass of solvent in kg
- M1= molar mass of solute
Elevation of boiling point
Depression of Freezing Point
The temperature at which the vapour pressure of the liquid present in a substance becomes equal to the vapour pressure of the solid present in the substance is known as the freezing point of the substance.
- Depression in freezing point can be defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.
- Let, T0f = boiling point of the pure solvent
- Tf = boiling point of the solution
- So the depression in freezing point can be given as -
ΔTf = T0f - Tf
- same as elevation in boiling point, depression in freezing point is directly related to molality ‘m’
ΔTf = 1000. Kf . m2 / M2 . m1
- Where,
- Kf= molal depression constant
- m2 = mass of solvent in g
- m1 = mass of solvent in kg
- M2= molar mass of solute
Depression of freezing point
Osmotic Pressure
The spontaneous flow of solvent molecules through a semipermeable membrane from a pure solvent to a solution is called osmosis. If extra pressure is applied from the solution, then it can stop the flow of the solvent molecules through the semipermeable membrane. This extra pressure is termed the Osmotic Pressure
- Experimentally, If Osmotic pressure =π and molarity = C
- Temperature = T
- Then,
- π ∝ CT [Also, π = CRT (where R is the gas constant)]
- π = (n2/ V) RT
Here,
- V = Volume of solution in litres
- n2 = moles of solute
- If m2 = weight of solute and M1 = Molar mass of solute
- then, n2 = m2 /M2
- Thus, π = m2RT/ VM
- Hence, by knowing the value of π, m2, T and V, the molar mass of the solute can be calculated.
Osmotic pressure
Ideal Solutions
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An ideal solution is a combination of substances in which the molecules of other species are different. Unlike the ideal gas, the molecules in the Ideal Solution use forces on one another.
- When these forces are equal for all molecules then a solution is said to be an ideal solution.
- An Ideal Solution obeys Raoult's law over the entire range of concentration.
Properties of Ideal Solution
Some important properties of an ideal solution are:
- The physical properties are strictly related to the properties of the pure components.
- The enthalpy of the solution is zero.
- The enthalpy of the solution gets quicker to zero; it is more likely to show an ideal behaviour. ΔmixH=0
- The volume of mixing is also zero. ΔmixV=0
Examples of Ideal SolutionsSome Ideal solution examples are:
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Non-Ideal Solutions
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When a solution does not obey Raoult’s law over the entire range of concentration and temperature ranges it is known as a non-ideal solution. A non-ideal solution may show positive or negative deviation from Raoult’s law.
- The vapour pressure of such a solution is either higher or lower than that predicted by Raoult's Law.
Properties of a Non-Ideal Solution
Some important properties of a non-ideal solution are:
- A non-ideal solution does not have any properties related to the properties of the pure components.
- The components do not obey Raoult’s law over the entire composition.
- The enthalpy of the solution is not equal to zero, ΔmixH≠0
- The volume of mixing is also not equal to zero. ΔmixV≠0
Examples of Non-Ideal SolutionsSome Non-Ideal solution examples are:
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Important Topics for JEE Main
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As per JEE Main 2024 Session 1, important topics included in the chapter Solutions are as follows:
Subtopics | Number of Questions Asked |
---|---|
Mole Concept | 5 |
Molarity and Molality | 2 |
Concentration of a Solution | 3 |
Raoult's Law | 1 |
Preparation of Standard Solution of Oxalic Acid | 1 |
Things to Remember
- Solution is a homogeneous mixture of a solute and solvent.
- It is of three types, namely solid, liquid and gaseous solutions.
- In this, the solute can dissolve solvent.
- Brass (an alloy of copper and zinc) is a common example of the solution.
- The properties of a solution upon the number of solute particles are known as colligative properties.
Previous Year Questions
- The elevation of the boiling point of the solution is 0.75K. The molecular weight of the solute in gmol−1 is
- 200mL of water is added to a 500mL of 0.2M solution. What is the molarity of this diluted solution?
- The mass of water in grams present in the solution is (Kf=1.86Kkgmol−1)
- what is the weight % and degree of dissociation (in %) of acetic acid in benzene?
- What is the molecular weight of the unknown solute?
- The degree of dissociation (α) of a weak electrolyte AxBy is related to van?t Hoff factor (i) by the expression
- If glucose of 36g weight is dissolved in 2kg of H2O then, change in boiling point (ΔTb) at 1.013 bar will be (Kb for H2O is 0.52Kkgmol−1)
- Volume of acid required to make 1 litre of 0.1MH2SO4 solution is:
- With increase in temperature, which one of these changes?
- Molarity is expressed as
Sample Questions
Ques. What is Raoult’s Law formula? (2 marks)
Ans. Raoult’s law states that the solution’s vapour pressure is the same as the sum of each volatile component’s vapour pressure if it was strictly multiplied by the mole fraction of that component in the solution. Raoult's is a chemical law that refers to a solution’s vapour temperature based on the mole fraction of a solution that is applied. Raoult’s law is expressed by the formula, P = P0 X.
Ques.Explain (A) Ideal gas (B) Supersaturated Solution? (2 marks)
Ans. (A) Ideal Gas: An ideal gas is clear as one where all collisions amid molecules or atoms are perfectly elastic and where there are no good-looking intermolecular forces. One can imagine it as a series of colliding perfectly hard spheres.
(B)Supersaturated Solution: A supersaturated solution can be depicted when a solute begins precipitating out after being dissolved at a specific concentration and temperature.
Ques. Why do gases become less soluble in liquids as the temperature is raised? (2 marks)
Ans. A release of heat energy, also called the exothermic reaction, is accompanied when gases are dissolved in water. With an increase in temperature, as per Lechatlier’s Principle, the equilibrium can be seen to move in the backward direction, causing gases to become less soluble in liquids.
Ques. Determine the mass of urea (NH2CONH2) which will be needed to make 2.5 kg of 0.25 molal aqueous solution? (3 marks)
Ans. As per the given question,
Moles of urea = 0.25 mole
Mass of solvent (NH2CONH2) = 60 g mol-1
Thus, it can be said that, 0.25 mole of urea = 0.25 x 60 = 15g
Mass of solution = 1000 + 15
= 1015 g
= 1.015 kg
1.015 kg of urea solution is known to have 15g of urea
\(\therefore\) 2.5 kg of solution urea =15/1.015 x 2.5
= 37 g
Ques. An aqueous solution with 2% non-volatile solute can be seen to exert a pressure of 1.004 bar at the boiling point of the solvent. Thus, calculate the molecular mass of the solute? (3 marks)
Ans. As per Raoult’s law, it can be said that:
Ques. Calculate the amount of CaCl2 (i = 2.47) that can be dissolved in a 2.5 litre of water in a way that its osmotic pressure becomes 0.75 atm at 27°C? (5 marks)
Ans. According to the given question, we can say,
π = i RT
⇒ π = i RT
Thus, as per the values given, we can say,
- π = 0.75 atm
- V = 2.5L
- i = 2.47
Hence,
T = (27 + 273)K = 300K
R = 0.0821 L atm K-1mol–
Molar mass of CaCl2(M) = 1 × 40 + 2 × 35.5 = 111 g mol -1
Finally,
w = \(\frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300} \)= 3.42g
Therefore, the needed amount of CaCl2 is 3.42g
Ques. Determine the mass of urea (NH2CONH2) which will be needed to make 3.5 kg of 0.25 molal aqueous solution? (3 marks)
Ans. As per the given question,
Moles of urea = 0.35 mole
Mass of solvent (NH2CONH2) = 60 g mol-1
Thus, it can be said that, 0.25 mole of urea = 0.35 x 60 = 21g
Mass of solution = 1000 + 21
= 1021 g
= 1.021 kg
1.021 kg of urea solution is known to have 15g of urea
\(\therefore\) 3.5 kg of solution urea =21/1.021 x 3.5
= 72 g
Ques. What is the difference between solute and solvent? (3 marks)
Ans. The difference between solute and solvent are as follows:
Parameter | Solute | Solvent |
---|---|---|
Definition | Solute is a substance which has the ability to dissolve. | Solvent is a dissolving medium. |
Boiling Point | Higher | Lower |
Physical State | Can be found in the solid, liquid or gas. | Mainly found in the liquid state, but can also be gaseous. |
Ques. Calculate the amount of CaCl2 (i = 2.47) that can be dissolved in a 3.5 litre of water in a way that its osmotic pressure becomes 0.85 atm at 27°C? (5 marks)
Ans. According to the given question, we can say,
π = i RT
⇒ π = i RT
Thus, as per the values given, we can say,
- π = 0.85 atm
- V = 3.5L
- i = 2.47
Hence,
T = (27 + 273)K = 300K
R = 0.0821 L atm K-1mol–
Molar mass of CaCl2(M) = 1 × 40 + 2 × 35.5 = 111 g mol -1
Finally,
w = (0.85 x 111 x 3.5 )/(2.47 x 0.0821 x 300) = 330.225 / 60.83
Therefore, the needed amount of CaCl2 is 5.42g
Ques. 2 ml of water is added to 6g of a powdered drug. The final volume is 3ml. Find the mass-by-volume percentage of the solution? (2 marks)
Ans. Given, Mass of solute = 6 g
- Volume of solution = 3 ml
- Mass by volume percentage
- (Mass of solute) / (volume of solution) × 100 %.
- (6 / 3) × 100 %
- 200%.
- Therefore, the mass by volume percentage is 200%.
Ques. Many people use a solution of NaPO, to clean walls before putting up wallpaper. The recommended concentration is 2.7% (m/v). Find the mass of NaPO needed to make 2.0L of the solution? (3 marks)
Ans. Given, Concentration of solution = 1.7% (m/v) [mass by volume percentage]
- Volume of Solution = 2 l = 2000 ml
- Mass by volume percentage = (Mass of solute) / (volume of solution) × 100
- 2.7 = (Mass of solute) / (2000) × 100
- Mass of solute = 54 g
- Therefore the mass required is 54g
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