CBSE Class 10 Syllabus physics
- Solution : A homogeneous mixture of two or more pure substances is known as solution.
- If the constituents of the solution are two, it is called binary, if three then ternary, if four then quaternary and so on.
- Two constituents of the solution are :
(i) Solute : A substance that is dissolved in another substance in lesser amount (a solvent), forming a solution. For example : Sugar, salt, etc.
(ii) Solvent : A substance in which another substance is dissolved in larger amount forming a solution. For example : Water, milk, etc.
Note: Solvent determines the physical state of the solution.
- Types of Solutions : Any state of matter (solid, liquid or gas) can act both as a solvent and as
a solute during the formation of a solution. Therefore, depending upon the physical states of
solute and solvent, we can have following nine different types of solutions :
| S. No. | Types of Solution | Solute | Solvent | Examples |
| 1. | Solid – Solid | Solid | Solid | Alloys like brass, bronze, etc. |
| 2. | Solid – Liquid | Solid | Liquid | Solution of sugar, salt, urea etc. in water. |
| 3. | Solid – Gas | Solid | Gas | Sublimation of substances like iodine, camphor, etc, into air, dust or smoke particles in air. |
| 4. | Liquid – Solid | Liquid | Solid | Hydrated salts, mercury in amalgamated zinc, etc. |
| 5. | Liquid – Liquid | Liquid | Liquid | Alcohol in water, benzene in toluene. |
| 6. | Liquid – Gas | Liquid | Gas | Aerosol, water vapour in air. |
| 7. | Gas – Solid | Gas | Solid | Hydrogen adsorbed in palladium. |
| 8. | Gas – Liquid | Gas | Liquid | Aerated drinks. |
| 9. | Gas – Gas | Gas | Gas | Mixture of gases, etc. |
- Aqueous solution : A solution containing water as solvent is known as aqueous solution. For example, sugar solution.
- Non-aqueous solution : A solution containing solvent other than water is known as non-aqueous solution. For example, iodine dissolved in alcohol.
- Saturated solution : A solution in which no more solute can be dissolved at the same temperature is known as saturated solution.
- Unsaturated solution : A solution in which more amount of solute can be dissolved at the same temperature is known as unsaturated solution.
- Solubility : Solubility can be defined as the maximum amount of solute that can be dissolved in 100 g of solvent to form a saturated solution at a given temperature.
- Causes of Solubility :
(i) Inter ionic attraction in the solute molecules : Molecules are stabilised in the lattice due to electrostatic forces and the energy released is known as lattice energy.
(ii) Inter molecular attraction between solvent molecules.
(iii) Solvation : It denotes the force of attraction between solute and solvent molecules.
(iv) Temperature.
- Factors affecting Solubility :
(i) Nature of Solute and Solvent : “Like dissolves like” i.e., polar solvents like water and ammonia can dissolve polar solute or ionic solute while non-polar solvents can dissolve non-polar organic solutes.
(ii) Temperature : Solubility increases with increase in temperature. It increases for endothermic reaction while it decreases for exothermic reaction.
(iii) Pressure : The solubility of solid in liquid is not affected significantly by pressure because solids and liquids cannot be compressed.
(iv) Hydration Energy : It is the amount of energy released when ions formed by 1 mole of ion get hydrated. It is an exothermic process.
- Method of expressing Concentration of Solution : The concentration of solution is the amount of solute present in the given quantity of solute or solvent. It can be expressed in any of the following types :
It is defined as mass of solute dissolved per 100 ml of solution. It is commonly used in medicine and pharmacy.
(iv) Parts per million (ppm) : It can be defined as the parts of a component per million (106) parts of the solution.
It is used to express the concentration of a solute present in trace quantities.
Parts per million can be expressed in three ways :
(v) Mole Fraction : It is the ratio of number of moles of a particular component to the total number of moles of all the components. e.g., mole fraction of component A.
where nA is the number of moles of component ‘A’ and nB is the number of moles of component ‘B’.
Similarly
Sum of mole fractions of all the components is always one.
χA + χB = 1
(vi) Molarity (M) : It is defined as the number of moles of solute per litre of solution.
where, WB = Weight of solute
V = Volume of solution in ml
MB = Molar mass of solute.
Unit is mol L–1 or M (molar).
(vii) Molality (m) : It is defined as the number of moles of solute per 1000 g or 1 kg of solvent.
where, WB = Weight of the solute
MB = Molar mass of solute
W = Mass of solvent in g
Unit is mol kg–1 or molal (m). Molality and mole fraction do not change with change in temperature.
(viii) Normality (N) : It is defined as number of gram equivalents of solute dissolved per litre of solution
where, WB = Mass of solute
EB = Equivalent weight of solute
V = Volume of solution in ml
- Relationship between Molarity (M) and Molality (m) :
where, m = Molality of solution
M = Molarity of solution
MB = Molar mass of solute
d = Density of solution in g ml–1
- Relationship between Mole fraction of solute (χB) and Molality (m) :
where χB is mole fraction of solute, m is molality and MA is molar mass of solvent.
- Henry’s Law : The relationship between pressure and solubility is guided by Henry’s Law. According to this law,
‘‘The mass of a gas dissolved in given volume of the liquid at a constant temperature depends upon the pressure applied.’’
It can also be stated as the partial pressure of the gas (p) in vapour phase is proportional to the mole fraction of the gas (χ) in the solution.
p = KHχ,
where KH = Henry’s constant.
begin{equation}
-frac{mathrm{d} theta}{mathrm{d} r} propto frac{1}{r^{2}} text { or } frac{mathrm{d} theta}{mathrm{d} r}=-frac{k}{r^{2}}
end{equation}
$-frac{mathrm{d} theta}{mathrm{d} r} propto frac{1}{r^{2}}$ or $frac{mathrm{d} theta}{mathrm{d} r}=-frac{k}{r^{2}}$
\(
-frac{mathrm{d} theta}{mathrm{d} r} propto frac{1}{r^{2}} text { or } frac{mathrm{d} theta}{mathrm{d} r}=-frac{k}{r^{2}}
\)
