Solutions - Topics, Books, FAQs

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A solution is a homogenous mixture that contains a solvent and a solute. A solvent is a substance which is present in a larger amount in solution and a solute is a substance which is present in a smaller amount. There are various common examples of solutions that you see in your daily life.

Other than these, there are several other examples as well which are very common for household purposes:

• Saltwater is formed when we mix salt in water.
• Vinegar is obtained when we mix acetic acid in water.
• A tincture of iodine is formed when we dissolve crystals of iodine in alcohol.
• Mouthwash consists of several chemicals dissolved in water.

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This Story also Contains

1. “A Clear Vision, Backed By Definite Plans, Gives You A Tremendous Feeling Of Confidence And Personal Power.” – Brian Tracy
2. Important Topics of Chapter Solution
3. Overview of Chapter
4. In this chapter, there are various important topics that you must understand completely:
5. How to Prepare for Solutions
6. Prescribed Books for Solutions

Important Topics of Chapter Solution

• Types of solutions
• Solubility
• Vapour pressure of liquid solutions
• Raoult's law
• Ideal and non-ideal solutions
• Colligative properties
• Abnormal molar mass

Overview of Chapter

There are various kinds of binary solutions exist in nature as given in the table below:

NCERT Notes Subject Wise Link:

• NCERT notes Class 11 Maths
• NCERT notes Class 11 Physics
• NCERT notes Class 11 Chemistry
• NCERT notes Class 11 Biology

In this chapter, there are various important topics that you must understand completely:

(i) Solubility: The maximum ability of a substance to get completely dissolved in a solvent. When the solute is solid and the solvent is liquid, then the only temperature affects the solubility. With the increase in temperature, the solubility increases. But pressure has no effect in this case as both solids and liquids are incompressible. But is the solute is gas, then pressure also is an important factor that plays a major role in insolubility. For this case, Henry's law is given which states that - "At constant temperature, the solubility of a gas is directly proportional to the partial pressure of a gas present above the solution".

(ii) Vapour pressure of liquid solutions: In this concept, you learn that the vapour pressure of volatile liquids in solution when taken in a closed vessel. This phenomenon is explained by Raoult's law, which states that - "The vapour pressure of each volatile liquid present in the solution is directly proportional to the mole fraction of that liquid present in solution".

Mathematically, Raoult's law can be expressed as follows:

ptotal =p10+(p20+p10)x2
where, the vapour pressure of pure component 1
p20 = vapour pressure of pure component 2
x₂ = mole fraction of component 2
(iii) Ideal and non-ideal solutions: Ideal solutions are those solutions which obey Raoult's law at all ranges of concentrations. Whereas the non-ideal solutions are those which do not obey Raoult's law at all ranges of concentrations. The vapour pressure of non-ideal solutions is always higher or lower than as predicted by Raoult's law and thus we say that the solution is exhibiting a positive or negative deviation, respectively.

(iv) Colligative properties: The properties of solutions which depend only on the number of solute particles present in the solution are known as colligative properties.

• Relative lowering of vapour pressure: The actual vapour pressure of a solvent in a solution is less than that of the pure solvent. Raoult explained this behaviour and established that this lowering of vapour pressure depends only on the concentration of the solute particles. Mathematically, it can be expressed as follows:Δp1p10=p10−p1p10=x2
• Elevation of the boiling point: The boiling point is the point of temperature when the vapour pressure of a substance is equal to the atmospheric pressure. This also depends upon the number of solute particles present in the solution.
  

Mathematically, it can be expressed as follows: ΔTb=Kbm
K_b = Boiling point elevation constant

Also Read:

• Depression of freezing point:

Freezing Point: It is the temperature at which the liquid and the solid form of the same substance are in equilibrium and have the same vapour pressure. A Solution freezes when its vapour pressure is equal to the V.P. of pure solid solvent. Due to the lower vapour pressure of the solution, the solid form of a solution separates out at a lower temperature.

ΔTf=Tf∘−Tf

• This is also termed cryoscopy and depression of freezing point (ΔT_f)
• It can be measured by Beckmann's thermometer method and Rast's method.
• For a dilute solution, ΔT_f is directly proportional to the molality (m) of the solution.
  Hence ΔT_f ∝ m
  ΔT_f = K_f m
  If the molality of the solution is one, then
  ΔT_f = K_f

  ΔTf and M can be found out by using these relations.

  ΔTf=KfwM×1000WM=Kf×w×1000ΔTf×W

  Here w= Weight of solute
  W= Weight of solvent
  Kf= Molal depression constant or cryoscopic constant
  M= Molar mass of Non-volatile solute.

  Kf=RT21000 Lf or ΔHfusion

  Here Lf or ΔHf= latent heat of fusion.

NOTE: The value of K_v or K_f depends on only the nature of the solvent and not on the nature of the solute.

NCERT Exemplar Solutions Subject-wise:

• NCERT exemplar solutions for class 11 Physics.
• NCERT exemplar solutions for class 11 Chemistry.
• NCERT exemplar solutions for class 11 Mathematics.
• NCERT exemplar solutions for class 11 Biology

Abnormal Mass: This concept says, that when a solute is dissolved in a liquid then the solute does not dissociate completely as expected instead it dimerizes and thus the molar mass of the solute becomes double. Such deviation of molar mass from the actual value is known as abnormal molar mass.

To deal with the case of an abnormal mass, a Vant Hoff factor was introduced in 1880, which mathematically is described as follows:
i= Normal Molar Mass Abnormal Molar Mass

How to Prepare for Solutions

• This chapter is a part of the physical chemistry and it is very important from the numerical point of view. The theory part is less but you simply need to work with the formulas and questions of this chapter.
• There are some common terms used in this chapter like vapour pressure and mole fraction, so before coming to this chapter, first, you must finish the two basic chapters - "Mole concept" and "Gaseous state".
• You must have completely memorized all the formulas and must have a very good grasp of solving the numerical problems.
• Usually, most of the students like this chapter as it is very easy and straightforward. This chapter holds a good weight ok marks in board exams as well as in competitive exams like JEE and NEET.

NCERT Solutions Subject-wise:

• NCERT solutions for class 11 Physics.
• NCERT solutions for class 11 Chemistry.
• NCERT solutions for class 11 Mathematics.
• NCERT solutions for class 11 Biology.

Prescribed Books for Solutions

For this chapter, first, you need to finish the important topics of solutions class 12 thoroughly from the class 12th NCERT book and then simultaneously solve the examples and questions given in the book. Apart from this, if you want to prepare for the advanced level for competitive exams like JEE and NEET, you must prepare from the books - O.P. Tandon and R.C Mukherjee. Meanwhile, in the preparation, you must continuously write the mock tests for the depth of knowledge. Our platform will help you to provide a variety of questions for deeper knowledge with the help of videos, articles and mock tests.