When a solution of haloalkane in either is treated with magnesium, we will get alkyl magnesium halide known as Grignard reagent, ethyl magnesium bromide is prepared from ethyl bromide as:
\(\underset { Ethyl \ bromide }{ { CH }_{ 3 }-{ CH }_{ 2 }Br+ } \)Mg \(\underrightarrow { dry \ ether } \)\(\underset { Ethyl \ magnesium \ bromide\\ (or)\\ Grignard \ reagent }{ { CH }_{ 3 }{ CH }_{ 2 }{ MgBr } } \)

₁H₂ or ₁D₂ . It occurs naturally in very small traces. It's nucleus consists of a proton, a neutron and one electron revolving around the nucleus. Its chemical properties are similar to those of protium but their reaction rates are different.

The property that is independent of the mass or the size of the system is called an intensive property.
Examples: Refractive index, Surface tension, density, temperature, Boiling point, Freezing point, molar volume, etc.,

The solubility product of NaCl is lower than that of NaOH. The more soluble a substance is, the higher the K_sp value it has In aqueous solution NaOH gives OH^- ions. It can be solvated by establishing H-bonds with water molecules. So it is more water soluble.

Mol.mass = 3(C) + 6(H) + 1(0)
= 3(12) + 6(1) + 1(16)
= 36 + 6 + 16 = 58

(i) l-butanol
(ii) I-chlorobutane
(iii) but-l-ene
(i) \(\underset { 1-buanol }{ CH_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }Cl } \overset { P/{ I }_{ 2 } }{ \longrightarrow } \underset { 1-Iodobutene }{ { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }I } \) 
(ii) \({ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }Cl+Nal\overset { Acetone }{ \longrightarrow } { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }I+Nacl\) 
(iii) 

(w) = 0.16 g
(x) = 0.35 g
\(\% S=\frac{32}{233} \times \frac{x}{\mathrm{w}} \times 100=\frac{32}{233} \times \frac{0.35}{0.16} \times 100=30.04 \%\)

Given : C = 0.25 M
T = 370.28 K
R = 0.0821 L atm mol^-1 K^-1
\(\pi\) = ?
\(\pi\) = CRT
= 0.25 x 0.0821 x 370.28
= 7.59 atm.

N₂O₂(g) \(\rightleftharpoons \) 2NO₂(g)
Kc = 0.21 at 373 K. The concentrations N₂O₄ and NO₂ are found to be 0.125 mol dm^-3 and 0.5 mol dm^-3 respectively at a given time. From the above information we can predict the direction of reaction as follows.
\(Q={[NO_2]^2\over [N_2O_4]}={0.5\times 0.5\over 0.125}=2\)
The Q value is greater than K_c. Hence, the reaction will proceed in the reverse direction until the Q value reaches 0.21.

The balanced equation for the formation of HCI is,
H₂(g) + CI₂(g) \(\rightarrow\) 2 HCI (g)
As per the stoichiometric equation, under given conditions,
To produce 2 moles of HCI, 1 mole of chlorine gas is required.
To produce 44.8 litres of HCI, 22.4 litres of chlorine gas are required.
\(\therefore\) To produce 11.2 litres of HCI,

= 5.6 litres of chlorine are required.

(i) Boron (Z = 5) ; 1s² 2s² 2p¹
(ii) Neon (Z = 10) ; 1s² 2S²2p⁶
(iii) Aluminium (Z = 13) ;1s² 2S²2p⁶ 3s² 3p¹
(iv) Chlorine(Z = 17) ; 1s² 2s² 2p⁶ 3s² 3p⁵
(v) Calcium (Z = 20) ;1s² 2S²2p⁶ 3s² 3p⁶ 4s²
(vi) Rubidium (Z = 37) ; 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹

(i) Isothermal process: Isothermal process is defined as one in which the temperature of the system remains constant, during the change from its initial to final states.
For an isothermal process dT = 0
(ii) Isobaric process: Isobaric process is defined as one in which the pressure of the system remains constant during its change from the initial to final state.
For an isobaric process dP = 0.
(iii) Isochoric process: Isochoric process is defined as one in which the volume of system remains constant during its change from initial to final state of the process.
For an isochoric processes dV = 0.

(i) Noble gases have completely filled electronic configuration and are stable.
(ii) It is difficult to remove the electrons from the valence shell. So the ionisation energy is maximum.

Alkali metals have one e^o in their outer most shell. Alkaline earth metals have 2 e^o in their outer most shell. More valence electrons and more positively charged nucleii leads to greater opportunity for metallic bonding.

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.

1. Gypsum is used in making drywalls or plaster boards.
2. Another important use of gypsum is the production of plaster of Paris. Gypsum is heated to about 300 degree Fahrenheit to produce plaster of paris, which is also known as gypsum plaster. It is mainly used as a sculpting material.
3. Gypsum is used in making surgical and orthopedic casts, such as surgical splints and casting moulds.
4. Gypsum plays an important role in agriculture as a soil additive, conditioner, and fertilizer. It helps loosen up compact or clay soil, and provides calcium and sulphur, which are essential for the healthy growth of a plant.
5. Gypsum is used in toothpastes, shampoos, and hair products.
6. Gypsum is a component of portland cement, where it acts as a hardening retarder to control the speed at which concrete sets.

All organic reactions can be understood by following the electron movements.
(i) Lone pair becomes a bonding pair.
(ii) Bonding pair becomes a lone pair.
(iii) A bond breaks and becomes another bond.
The electron movement depends on the nature of the substrate, reagent and the prevailing conditions.
Type 1.A lone pair to a bonding pair

Type 2. A bonding pair to a lone pair

Type 3. A bonding pair to an another bonding pair

(i) CH₃-CH = CH₂ + H₂SO₄ \(\rightarrow\) \({ CH }_{ 3 }-\underset { \overset { | }{ OSO_{ 2 } } OH }{ CH } -{ CH }_{ 3 }\overset { { H }_{ 2 }O }{ \longrightarrow } \) \({ CH }_{ 3 }-\underset { \overset { | }{ OH } }{ CH-{ CH }_{ 3 } } +{ H }_{ 2 }{ SO }_{ 4 }\)
                                    2-Propyl hydrogen sulphate       2-proponel

(ii) 

(iii) CH₃ - CH = CH₂ + HBr \(\overset { Peroxide }{ \underset { ({ C }_{ 6 }{ H }_{ 5 }CO)_{ 2 }{ O }_{ 2 } }{ \longrightarrow } } \) CH₃ - CH₂ - CH₂ - Br
                                                               1-Bromopropane
(iv) nCH₃ - CH = CH₂ \(\overset { Reol\ Hot\ Fe\ Tube }{ \underset { 873\ K }{ \longrightarrow } } \)\({ \left[ \underset { \overset { | }{ { CH }_{ 3 } } }{ \sim \sim CH } -CH_{ 2 }\sim \sim \right] }_{ n }\)
                                                           Polypropene
(v) \({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ C } ={ CH }_{ 2 }\overset { KMn{ O }_{ 4 }/{ H }^{ + } }{ \underset { -{ CO }_{ 2 },-{ H }_{ 2 }O }{ \longrightarrow } } { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ C } =O\)
2-methyl prop-1-ene                            Propan-2-one

The nature of the deviation from the Rauolt's law can be explained in terms of the intermolecular interactions between solute (A) and solvent (B). Consider a case in which the intermolecular attractive forces between A and B are weaker than those between the molecules of A (A - A) and molecules of B (B - B). The molecules present in such a solution have a greater tendency to escape from the solution when compared to the ideal solution formed by A and B, in which the intermolecular attractive forces (A - A, B - B, A - B) are almost similar. Consequently, the vapour pressure of such non-ideal solution increases and it is greater than the sum of the vapour pressure of A and B as predicted by the Raoult's law. This type of deviation is called positive deviation.
Here,  \({ p }_{ A }>{ p }_{ A }^{ 0 }{ x }_{ A }\) and \({ p }_{ B }>{ p }_{ A }{ x }_{ B }\)
Hence \({ p }_{ total }>{ p }_{ A }^{ 0 }{ x }_{ A }+{ p }_{ B }^{ 0 }{ x }_{ B }\)
Letus understand the positive deviation by considering a solution of ethyl alcohol and water. In this solution the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alcohol and water-water interactions). This results in the increased evaporation of both components from the aqueous solution of ethanol. Consequently, the vapour pressure of the solution is greater than the vapour pressure predicted by Raoult's law. Here, the mixing process is endothermic i.e. \(\triangle\)H_mixing > 0 and there will be a slight increase in volume (\(\triangle\)V_mixing > 0).
Examples for non - ideal solutions showing postive deviations: 
Ethyl alcohol & cyclohexane, Benzene & acetone, Carbon tetrachloride & chloroform, Acetone & ethyl alcohol, Ethyl alcohol & water.

\(\therefore\) The empirical formula is C₉H₈O₄.

According to charles law
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { 1.54 }{ 293 } \)
= 0.0052
T₁ = 20 C + 273 = 293 K
T₂ = 40°C
\(\frac { { V }_{ 2 } }{ { T }_{ 2 } } =\frac { 1.65 }{ 40+273 } =\frac { 1.65 }{ 313 } =0.0052\)
T₃ = 100°C + 273 = 373 K
\(\frac { { V }_{ 3 } }{ { T }_{ 3 } } =\frac { 1.95 }{ 373K } =0.0052\)
T₄ = 120°C+273 = 393 K
\(\frac { { V }_{ 4 } }{ { T }_{ 4 } } =\frac { 2.07 }{ 393 } =0.0052\)
The average value of constant of proportionality is 0.0052.

2H₂ + O₂ ➝ 2H₂O
The isotope of hydrogen which has one proton and one neutron in its nucleus is Deuterium.
2D₂ + O₂ ➝ 2D₂O
The product is heavy water (Deuterium oxide). H₂O and D₂O have same chemical properties but the
reaction velocity of D2O is slightly less due to the difference in the mass number of the isotopes known as isotopic effect. Deuterium is heavier than protium so reacts slowly.

Let us use the Born - Haber cycle for determining the lattice enthalpy of NaCl as follows:
Since the reaction is carried out with reactants in elemental forms and products in their standard states, at 1 bar, the overall enthalpy change of the reaction is also the enthalpy of formation for NaCl. Also, the formation of NaC1 can be considered in 5 steps. The sum of the enthalpy changes of these steps is equal .to the enthalpy change for the overall reaction from which the lattice enthalpy of NaCl is calculated.
Let us calculate the lattice energy of sodium chloride using Born-Haber cycle

Δ°H_f = heat of formation of sodium chloride 
= - 411.3 kJ mol^-1
Δ°H₁= heat of sublimation ofNa(S) = 108.7 kJ mol^-1
Δ°H₂ = ionisation energy ofNa(S) = 495.0 kJ mol^-1
Δ°H₃= dissociation energy ofCI2(S) = 244 kJ  mol^-1
Δ°H₄= Electron affinity ofCl(S) = - 349.0 kJ mol^-1
Δ°H_f = Δ°H₁+Δ°H₂+1/2Δ°H₃+Δ°H₄+Δ°H₅
Δ°H₅=(Δ°H_f )-( Δ°H₁+Δ°H₂+1/2Δ°H₃+Δ°H₄)
⇒Δ°H₅ = (-411.3)-(108.7+495.0+ 122-349)
Δ°H₅= (-411.3)-(376.7)
∴ Δ°H₅ = -788 kJ mol^-1
This negative sign in lattice energy indicates that the energy is released when sodium is formed from its constituent gaseous ions Na⁺ and Cl^-

m 160 g = 160 x 10^-3 kg
\(v=140 km hr^{-1}=\frac{104\times10^{3}}{60\times60}ms^{-1}\)
\(v=38.88ms^{-1}\)
\(\lambda=\frac{h}{mv}\)
\(=\frac{6.626\times^{-34}Kgm^{2}s^{-1}}{160\times10^{-3}Kg\times38.88ms^{-1}}\)
\(\lambda=1.065\times10^{-34}m\)

r(K⁺)+r(Cl^-) = d(K^+-Cl-) = 3.14 Å.
The effective nuclear charge for K+ and CI- can be calculated as follows.
K+ = (1s²)          (2s²2p⁶)        (3s²3p⁶)
        inner shell    (n-1)th shell   n^th shell
Z*(K-) = Z-S
= 19 - [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 19 - 11.25 = 7.75
Z*(Cl-) = 17- [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 17-11.25 = 5.75
∴ \(\frac { r({ K }^{ + }) }{ r(Cl^{ - }) } =\frac { Z*(Cl^{ - }) }{ Z*(K^{ + }) } =\frac { 5.75 }{ 7.75 } \)=0.74
∴ r(K⁺) = 0.74 r(Cl^-)
Substitute (2) in (1)
0.74 r(Cl^-) + r(Cl^-) = 3.14 Å.
1.74 r(Cl^-) = 3.14 Å
r(Cl^-) = \(\frac { 3.14\overset { 0 }{ A } }{ 1.74 } \)=1.81.Å.