Ideal solutions: The solutions which obey Raoult's law over the entire range of concentration are known as ideal solutions. Ideal solutions are formed by mixing the two components which are identical in molecular size, in structure and have almost identical intermolecular forces.
Examples:
(i) Benzene and toluene
(ii) n-Hexane and n-Heptane
(iii) Chlorobenzene and bromobenzene.
Characteristics:
(i) They must obey Raoult's law.
(ii) ΔH mixing should be zero.
(iii)ΔV mixing should be zero, i.e. volume change on mixing is zero.

Non-ideal solutions : The solutions which do not obey Raoult's law are called non-ideal solutions. In case of non-ideal solutions there is change in volume and heat energy when the two componnts are mixed.
Characterstics :
(i) They does not Rault's law.
(ii)△V = mix ≠ 0
(iii) △H mix ≠ 0
Behaviour of Ideal Solutions: A plot of P_I or P₂ versus the mole fraction x₁ and x2 for an ideal solution gives a linear plot. These Lines (I and II) pass through the points and respectively when x _I and x₂ is equal to unity. Similarly the plot (Line III) of P_total versus x₂ is also linear. The minimum value of P_total is P_total and the maximum value is P₂⁰, assuming that component I is less volatile than component 2, i.e. P1 O < P2 O

Let us consider the formation of HI in which, 'a' moles of hydrogen and 'b' moles of iodine gas are allowed to react in a container of volume V. Let 'x' moles of each of H₂ and I₂react together to form 2x moles of HI.
\({ H }_{ 2\left( g \right) }+{ I }_{ 2\left( g \right) }\rightleftharpoons { 2HI }_{ \left( g \right) }\)

Applying law of mass action,
\({ K }_{ C }=\cfrac { { \left[ HI \right] }^{ 2 } }{ { \left[ H \right] }_{ 2 }\left[ { I }_{ 2 } \right] } \)
= \(\cfrac { \left( \cfrac { 2x }{ V } \right) ^{ 2 } }{ \left( \cfrac { a-x }{ V } \right) \left( \cfrac { b-x }{ v } \right) } =\cfrac { { 4x }^{ 2 } }{ \left( a-x \right) \left( b-x \right) } \)
The equilibrium constant Kp can also be calcu•lated as follows:
We know }he \rer~tionship between the K_c and K_p
Here the \(\Delta n_{ g }\)=n_p -n_r = 2 - 2 =0
Hence K = K_c ;\({ K }_{ p }=\cfrac { { 4x }^{ 2 } }{ \left( a-x \right) \left( b-x \right) } \)

(i) Bond length (d_H-H) in hydrogen molecule = 0.74A
 \(\therefore\) Covalent radius of hydrogen = \(\frac { d_{ H-H } }{ 2 } -\frac { 0.74 }{ 2 } =.37A\)
(ii) Bond length (d_CI-Cl) in chlorine molecule = 1.98A
\(\therefore\) Covalent radius of chlorine = \(\frac { d_{ cl-CL } }{ 2 } =\frac { 1.98 }{ 2 } =0.99A\)
(iii) Bond length (C - C) in diamond = 1.54A
\(\therefore\)  Covalent radius of carbon = \(\frac { 1.54 }{ 2 } =0.77A\)
(iv) Bond length of (d_si-c) in carborundum = 1.93A
\({ d }_{ si-c }={ r }_{ si }+{ r }_{ c }\)
\(1.93\quad =\quad { r }_{ si }+0.77\)Refer (iii)
\({ r }_{ si }=1.93-0.77=1.16A\)
(v) Bond length (d_C-CI) in CCl₄ = 1.76A
\({ d }_{ c-cl }={ d }_{ c }+{ d }_{ cl }\)
1.76 \(=r_{ c }+0.99\) Refer (ii)
\({ r }_{ c }=1.76-0.99=0.77A\)

(i) Quick lime :
Preparation :
It is produced on a commercial scale by heating limestone in a lime kiln at 1173K (1070-1270).
\(CaCO_3\leftrightharpoons CaO + CO_2\)
Uses: Calcium oxide is used
(i) to manufacture cement, mortar and glass.
(ii) in the manufacture of sodium carbonate and slaked lime.
(iii) in the purification of sugar.
(iv) as drying agent
(ii) Slaked lime :
Preparation: Calcium hydroxide is prepared by adding water to quick lime, CaO.
CaO + H₂O ⟶ Ca(OH)₂
Uses: Calcium hydroxide is used
(I) in the preparation of mortar, a building material.
(ii) in white wash due to its disinfectant nature.
(iii) in glass making, in tanning industry, for the preparation of bleaching powder and for purification of sugar

q = 957.7 kJ
T = 110+273 = 383 K
ΔS= \(\frac { { q }_{ rev } }{ T } \)
ΔS = \(\frac { 957.7 }{ 383 } \)
ΔS =2.5 kJ K^-1

n = 5; V = 2 Iitre;T= 27 + 273 = 300 K
\(P=\frac { nRT }{ V } [\because PV=nRT]\)
\(P=\frac { 5\times 8.314\times 300 }{ 2 } =62.355\quad bar\)
\(\therefore\) The pressure of CI₂ gas will be 62.355 bar.

(i) Water molecules form strong hydrogen bonds with one another. For example, each water molecule is linked to four others through hydrogen bonds The strong hydrogen bonding prevails over a short range and therefore the denser packing in water.
(ii) In ice, each atom is surrounded tetrahedrally by four water molecules through hydrogen bonds. That is, the presence of two hydrogen atoms and two lone pairs of electron on oxygen atoms,in each water molecule allows formation of a three-dimensional structure. This arrangement creates an open structure, which accounts for the lower density of ice compared with water at 0^oC.

Decomposition reaction: Redox reactions in which a compound breaks down into two or more components are called decomposition reactions. These reactions are opposite to combination reactions. In these reactions, the oxidation number of the different elements in the same substance is changed.

Limitation of Bohr's atom model:
(a) The Bohr's atom model is applicable only to species having one electron such as hydrogen, Li²⁺ etc ... and not applicable to multi electron atoms.
(b) It was unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).
(c) Bohr's theory was unable to explain why the electron is restricted to revolve around the nucleus in a fixed orbit in which the angular momentum of the electron is equal to nh/2π