Given: M = 12 M;
d = 1.2 gL^-1
In 12 M - HCI means 12 mole of HCI in 1 litre of the solution (i.e)
n = 12 mole
Calculation of mass of 1L of HCI solution:
Mass of 1 L of HCl solution = d x v
= 1.2 x 1000 = 1200 g
Mass of HCI (m) = No. of moles of HCl x Molar mass of HCl
= nm
= 12 x 36.5
= 438 g
.'. Mass of water (solvent) = 1200 - 438
= 762 g = 762 x 10^-3 kg
\(\therefore \text { Molality }=\frac{\text { No.of moles of solute }}{\text { Mass of solvent }(\mathrm{kg})}\)
\(=\frac{12}{762 \times 10^{-3}}\)
= 15.75 m.

2 NO(g) + O₂ (g) ⇌ 2NO₂(g)

\(K_p={P^2_{NO_2}\over P^2_{NO_2}.Po_2}\)
\(={0.96\times 0.96\over 0.04\times 0.04\times 0.52}\)
= 11.07 x 10² (atm)^-1
K_eq = 41.6 x 10² M^-1.

For the process, H₂O(s) \(\rightleftharpoons \) H₂O(l)
\(\triangle\)S(fusion) = \(\frac { \Delta H(fusion) }{ Freezingtemperature } \)
= \(\frac { 6\times 1000 }{ 273 } =21.98\) JK^-1 mol^-1

For a fixed mass of a gas at constant pressure, the volume is directly proportional to its temperature (K). Mathematically it can be represented as (at constant P and n)
or  V = kT
or \(\frac { V }{ T } =Constant\)
Use of hot air balloon in meteorological observation is based on Charles law.

a) Electron-precise hydrides:
These have required number of electron to represent their conventional d lewis structure. All the elements of carbon group (14) form such hydrides.
Eg CH₄, C₂H₆, SiH₄, GeH₄
b) Electron - deficient hydrides:
There act as Lewis acids (ie) electron acceptors. The elements from group (13) form such hydrides.
Eg: B₂H₆
c) Electron - rich hydrides:
There have excess electron, which are present as lone pairs. Elements of group 15-17 form such hydrides. They behave as Lewis bases (ie) electron donors.
Eg: NH₃ ,H₂O, HF

i) chiral molecule:
The chiral molecule is
ii) S_N2 reaction:
undergoes S_N2 reaction faster as it is a primary alkyl halide.
\( \text { (or) } \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl}\\ 1 - chlorobutane\)

Benzene undergoes nitration reaction becasuse it is an elecron - rich system due to delocalised \(\pi\) electron. So it is easily attacked by electrophiles and gives substituted products.

Mechanism:
Step - I : Formafion of the electrophile:

Step - II : Formation of carbocation intermediate:
The electrophile attacks the aromatic ring to form a carbocation intermediate which is stabilised by resonance.

Step - III: Formation of nitro benzene:
The electrophile attacks the artomatic ring to fo

Let the volume of 4M HCl required to prepare 500 mL of 2.5 MHCl = x mL
Therefore, the required volume of 2M HCl = (500 - x) mL
We know from the equation
C₁V₁+ C₂V₂ = C₃V₃
(4x) + 2(500 - x) = 2.5 x 500
4x + 1000 - 2x = 1250
2x = 1250 - 1000
\(x={250\over 2}\)
= 125 ml
Hence, volume of 4M HCl required = 125 mL
Volume of 2M HCl required = (500 - 125) mL = 375 mL.

\(K_c={[AB]^2\over [A_2][B_2]}\)
A - green
B - blue

Given that 'V' is constant (closed system)
At equilibrium
\(K_c={({4\over V})^2\over ({2\over V})({2\over V})}={16\over 4}=4\)
K_P = K_C (RT)^Δn
K_P = 4(RT)^o = 4
At Stage 'x'
\(Q={({6\over V})^2\over ({2\over V})({1\over V})}={36\over 2}=18\)
Q > K_C ie., reverse reaction is favoured
At Stage 'y'
\(Q={({3\over V})^2\over ({3\over V})({3\over V})}={9\over 3\times 3}=1\)
K_C > Q ie., forward reaction is favoured.

i) \(K_{p}=K_{c}(R T)^{\Delta n_{g}}\)
K_p = Equilibrium constant in term of partial Pressures.
K_c = Equilibrium constant in term of concentration.
R = Gas constant; T = Temperature
\(\Delta \mathrm{n}_{\mathrm{g}}\) = Difference between the sum of number of moles of products and the sum of number of moles of reactants in gas phases.
ii) Synthesis of HI:
\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\Delta n_{g}=0 \therefore K_{p}=K_{c}(R T) \Delta n_{g}\)
\(K_{p}=K_{c}(R T)^{\circ} \)
\(K_{p}=K_{c} \text {. }\)

The shape of an atomic orbital is found by finding the probability (\({ \Psi }^{ 2 }\) ) of the electron at different points around the nucleus and representing the density of the nucleus. Same orbitals are found to have a region of space where the probability of finding electron is zero. This is called a node.
(i) All 's' orbitals are spherical in shape.
(ii) All 'p' orbitals are dumb-bell in shape.
The three 'p' orbitals differ in orientation. They lie along the axes and called P_x, P_y and P_z orbitals.
(iii) All 'd' orbitals have clover- leaf shaped with four lobes. Three have lobes between axes and are called d_xy, d_yz, d_xz orbitals.
The fourth lobe is along the axes and is called \({ d }_{ { x }^{ 2 }-{ y }^{ 2 } }\) the fifth has two lobes along 'z' axis and dough-nut shape in the centre and is called \({ d }_{ { z }^{ 2 } }\)

\({ O }_{ (g) }+{ e }^{ - }\longrightarrow { O }^{ - }_{ (g) };{ \Delta }_{ eg }H=-141\quad kj\quad { mol }^{ -1 }\)
\({ O }_{ (g) }+{ e }^{ - }\longrightarrow { O }^{ 2- }_{ (g) };{ \Delta }_{ eg }H=+780\quad kj\quad { mol }^{ -1 }\)
When an electron is added to oxygen atom to form ion, energy is released. Hence, first electron gain enthalpy of oxygen is negative.
But when another electron is added to O^- ion to form O^2- ion, it feels stronger electrostatic repulsion. Hence, addition of second electron takes place with absorption of energy. That's why the second electron gain enthalpy of oxygen is positive.

The reaction is
N₂ + 3H₂ \(\longrightarrow\) 2NH₃
1mol  3mol       2 mol
2 x 14 = 2(1 x 14 + 3 x 1)
28 g = 34 g
Thus, to produce 34.0 g ammonia, 28 g of N₂ is required

\({ He }^{ + }\longrightarrow { He }^{ 2+ }+{ e }^{ - }\)
\({ E }_{ n }=\frac{-13.6(2)^2}{n^2}\)
\({ E }_{ 1 }=\frac{-13.6(2)^2}{(1)^2}=-54.4\)
\({ E }_{ \infty }=\frac{-13.6(2)^2}{(\infty)^2}=0\)
\(\therefore\) Required Energy for the given process
= E_\(\infty\) - E_I = 0 - (- 54.4) = 56.4 ev.

Electronic configuration of oxygen

ஃ 8^th electron present in 2p_x orbital and the quantum numbers are
n = 2,l = 1,m₁ = either + 1 or -1 and s = -1/2
Electronic configuration of chlorine 

15^th electron present in 3P_z orbital and the quantum numbers are n = 3, l = 1, m₁ = either +1 or -1 and ms = +1/2

It will be more difficult to drink water with a straw on the top of a Mount Everest. This is because the reduced atmospheric pressure is less effective in pushing water up into the straw, below the water surface.
The force that propels water through a straw is atmospheric pressure, which is less at high altitude.

Inductive effect (I):
(i) Inductive effect is defined as the change in the polarisation of a covalent bond due to the presence of adjacent bonds, atoms or groups in the molecule. This is a permanent phenomenon.
(ii) Let us explain the inductive effect by considering ethylchloride as example. The C-C bond in ethyl chloride is polar.
We know that chlorine is more electronegative than carbon, and hence it attracts the shared pair of electron between C-Cl in ethyl chloride towards itself. is develops a slight negative charge on Chlorine and a slight positive charge on carbon to which chlorine is attached.
To compensate it, the C₁ draws the shared pair of electron between itself and. C₂ . This polarisation effect is called inductive effect.
(iii) The magnitude of the charge separation decreases rapidly, as we move away from C₁ and is observed maximum for 2 carbons and almost insignicant after 4 bonds from the active group.
\(\overset { \delta }{ C } \overset { \delta + }{ { H }_{ 3 } } \longrightarrow \underset { 1 }{ C\overset { \delta + }{ { H }_{ 2 } } } \twoheadrightarrow \overset { \delta - }{ C } { l }_{ 2 }\)
It is important to note that the inductive effect does not transfer electrons from one atom to another but the displacement effect is permanent. The inductive effect represents the ability of a particular atom or a group to either withdraw or donate electron density to the attached carbon. Based on this ability the substituents are classified as +I groups and -I groups. Their ability to release or withdraw the electron through sigma covalent bond is called +I effect and -I effect respectively.
Highly electronegative atoms and atoms of groups which are carry a positive charge are electron withdrawing or -I group
Example : -F, -CI, -COOH, -NO₂, NH₂,
Higher the electronegativity of the substitutent, greater is the -I effect.
The order of the -I effect of some groups are given below :
NH³⁺ > NO₂ > CN > SO₃H > CHO > CO > COOH > COCI > CONH₂ > F > Cl > Br > I > OH > OR, NH₂ > C₆H₅ > H
Highly electropositive atoms and atoms are groups which carry a negative charge are electron donating or +I groups.
Example. Alkali metals, alkyl groups such as methyl, ethyl, negatively charged groups such as CH₃O^-, C₂H₅O^-, COO^- etc.
Lesser the electronegativity of the elements, greater is the +I effect. The relative order of +I effect of some alkyl groups is given below
\(-\mathrm{C}\left(\mathrm{CH}_3\right)_3>-\mathrm{CH}\left(\mathrm{CH}_3\right)_2>-\mathrm{CH}_2 \mathrm{CH}_3>-\mathrm{CH}_3\)
Let us understand the influence of inductive effect on some properties of organic compounds.
Reactivity :
When a highly electronegative atom such as halogen is attached to a carbon then it makes the C-X bond polar. In such cases the -I effect of halogen facilitates the attack of an incoming nucleophile at the polarised carbon, and hence increases the reactivity.

If a - I group is attached nearer to a carbonyl carbon, it decreases the availability of electron density-on the carbonyl carbon, and hence increases the rate of the nucelophilic addition reaction.
Aciditv of carborvlic acids :
When a halogen atom is attached to the carbon which is nearer to the carboxylic acid group, its -I effect withdraws the bonded electrons towards itself and makes the ionisation of H⁺ easy. The acidity of various chloro acetic acid is in the following order. The strength of the acid increases with increase in the -I effect of the group attached to the carboxyl group.
Tiichloro acetic acid > Dichloro acetic acid > Chloro acetic acid > acetic acid

Homologous series: A series of organic compounds each containing a characteric functional group and the successive' members differ from each other in molecular formula by a CH₂ group is called homologous series. Eg.
Alkanes : Methane (CH₄), Ethane (C₂H₆), Propane (C₃Hg) etc .
Alcohols: Methanol (CH₃OH), Ethanol (C₂H₅OH) Propanol (C₃H₇OH) etc ..)
Compounds of the homologous series are represented by a general formula Alkanes CnH_2n+2' Alkenes C_nH_2n, Alkynes C_nH_2n-2 and can be prepared by general methods. They show regular gradation in physical properties but have almost similar chemical property.

Mass of He = 0.4 g; V = 10 Litres
Mass of O = 1.6 g; T = 27^oC + 273 = 300K
Mass of N = 1.4g
R = 0.0821 dm³ atm K^-1 mol^-1
Sol:
Calculate of no.of moles:
\(\left[ No.of\ moles=\frac { mass }{ molecular\ mass } \right] \)
No.of moles of He (n_He) = \(\frac { 0.4 }{ 4 } =0.1\)
No.of moles O₂ (\({ n }_{ { o }_{ 2 } }\)) = \(\frac { 1.6 }{ 32 } =0.05\)
No. of moles of N₂ (\({ n }_{ { N }_{ 2 } }\)) =\(\frac { 1.4 }{ 28 } =0.05\)
Total no.of moles (n) = n_He + \({ n }_{ { o }_{ 2 } }\) + \({ n }_{ { N }_{ 2 } }\)
= 0.1 + 0.05 + 0.05
n = 0.2
Calculate of mole fraction:
Mole fraction of He (X_He) = \(\frac { 0.1 }{ 0.2 } =0.5\)
Mole fraction of O₂  (\({ X }_{ { o }_{ 2 } }\)) =  \(\frac { 0.05 }{ 0.2 } =0.25\)
Mole fraction of N₂  ( \({ X }_{ { N }_{ 2 } }\)) = \(\frac { 0.05 }{ 0.2 } =0.25\)
Calculate of Total pressure of an ideal gas:
PV = nRT
\(P=\frac { nRT }{ V } \)
\(=\frac { 0.2\times 0.082\times 300 }{ 10 } =0.4926\)
P = 0.4926 atm
Calculation of Partial pressure:
Partial pressure of He (P_He) = 0.5 x 0.4926
= 0.2463 atm
[\(\therefore\) Partial pressure = mole fraction x Total pressure]
Partial pressure of O₂ (\({ P }_{ { o }_{ 2 } }\)) = 0.25 x 0.4926 = 0.1231 atm
Partial pressure of N₂ (\({ P }_{ { N }_{ 2 } }\)) = 0.25 x 0.4926 = 0.1231 atm
Calculation of Total pressure in the cylinder:
\(\therefore\) Total pressure P_Total = P_He + \({ P }_{ { o }_{ 2 } }\) + \({ P }_{ { N }_{ 2 } }\)
= 0.2463 + 0.1231 + 0.1231
= 0.4925
P_Total = 0.4925 atm

Born - Haber cycle for the formation of CaCl₂
Born - Haber cycle is used to calculate the lattice enthalpy of CaCl₂ 

Δ^oH₁ - Enthalpy change for the sublimation of Ca_(s) to Ca_(g)
Δ^oH₂ - Enthalpy change for dissociation of Cl_2(g) to 2Cl_(g)
Δ^oH₃ - Ionisation energy for Ca_(g) to Ca²⁺_(g)
Δ^oH₄ - Electron affinity for the conversion of 2Cl_(g) to 2Cl^-_(g)
Δ^oH₅ - Lattice enthalpy for the formation of solid CaCl₂·
Δ^oH_f =Δ^oH₁+Δ^oH₂+Δ^oH₃+Δ^oH₄+Δ^oH₅

Screening effect: The repulsive force between the inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus, the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called shielding effect.
Pauling's scale: Pauling, he assigned arbitrary value of electronegativities for hydrogen and fluorine as 2.2 and 4.0 respectively. Based on this the electronegativity values for other elements can be calculated using the following expression.
\(({ X }_{ A }-{ X }_{ B })=0.182\sqrt { E_{ AB } } -({ E }_{ AA }*{ E }_{ BB })^{ 1/2 }\)
Where E_AB' EAA and E_BB are the bond dissociation energies of AB, A₂ and B₂ molecules respectively. The electronegativity of any given element is not a constant and its value depends on the element to which it is covalently bound. The electronegativity values play an important role in predicting the nature of the bond.