(Λ^o_HoAC) = [(Λ^o)_HCl + Λ^o)_NaOAC ] - (Λ^o)_NaCl
= (426.2 + 91) - (126.5)
= 390.7

BaSO₄ ⇌ Ba²⁺ + SO₄^2-
K_sp = (s) (s)
K_sp = (s)²
= (2.42 × 10^-3gL^-1)²
\(=\frac{2.42 \times 10^{-3} gL^{-1}}{233 \text{ g mol}^{-1}}\)
= (0.01038 x 10^-3)²
= (1.038 x  x 10^-5)²
= 1.077 x 10^-10
= 1.08 × 10^-10mol²L^-2

rate₁ = k[0.1]ⁿ [0.1]^m ....(1)
rate₂ = k[0.2]ⁿ [0.1]^m ....(2)
(2) (1)
\(=\frac{2x}{x} = \frac{k[0.2]^n [0.1]^m}{k[0.1]^n [0.1]^m}\)
\(= \frac{2x}{x} = 2^n\)
n = 1
rate₃ = k[0.1]ⁿ [0.2]^m ....(3)
rate₄ = k[0.2]ⁿ [0.2]^m ....(4)
\(=\frac{8x}{2x} = \frac{k[0.2]^n [0.2]^m}{k[0.2]^n [0.1]^m}\)
= 8/2 = 2m
m = 2
rate = k[A] [B]²

\(\rho=\frac{\mathrm{n} \times \mathrm{M}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}}\)
For bcc n =2, M = 39 nearest distance 2r = 4.52 A⁰
\(a =\frac{4r}{\sqrt {3}} = \frac { 2\times 4.52 \times 10^{-10} }{ \sqrt{3}} \)
\(\rho =\frac { {( 5.21\times 10^{-10})^{ 3 }} \times (6.023 \times 10^{-23}) }{ 2 \times 39} \)
ρ = 915 kgm^-3

(a) coordination isomers
(b) no isomerism ( different molecular formula)
(c)⬅NCS; ⬅SCN coordinating atom differs : linkage isomers

As we move from La to Lu, their metallic behaviour because almost similar to that of aluminium.

Cr ⇒ [Ar]3d⁵4s¹

Structure of fructose: Fructose is the sweetest of all known sugars. It is readily soluble in water. Fresh solution of fructose has a specific rotation -1330 which changes to - 920 at equilibrium due to mutarotation. Similar to glucose, the structure of fructose is deduced from the following facts.
(i) Elemental analysis and molecular weight determination of fructose show that it has the molecular formula C₆H₁₂O₆.
(ii) Fructose on reduction with HI and red phosphorus gives a mixture of n - hexane (major product) and 2 - iodohexane (minor product). This reaction indicates that the six carbon atoms in fructose are in a straight chain.

(ii) Fructose reacts with NH₂OH and HCN. It shows the presence of carbonyl groups in the molecule of fructose.
(iv) Fructose reacts with acetic anhydride in the presence of pyridine to form penta acetate. This reaction indicates the presence of five hydroxyl groups in a fructose molecule. ,
(v) Fructose is not oxidized by bromine water. This rules out the possibility of presence of an aldehyde (-CHO) group.
(vi) Partial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol and mannitol which are epimers at second carbon. New asymmetric carbon is formed at C-2. This confirms the presence of keto group.

(vii) On oxidation with nitric acid, it gives glycolic acid and tartaric acids which contain smaller number of carbon atoms than in fructose.

This shows that a keto group is present in C-2. It also shows the presence of 10 alcoholic groups at C- 1 and C- 6

The monomers are phenol and formaldehyde. The polymer is obtained by the condensation polymerization of these monomers in presence of either an acid or a base catalyst.
Phenol reacts with methanal to form ortho or para hydroxyl methylphenols which on further reaction with phenol gives linear polymer called novolac. Novalac on further heating with formaldehyde undergo cross linkages to form Bakelite.

Uses:
Navolac is used in paints. Soft bakelites are used for making glue for binding laminated wooden planks and in varinishes, Hard bakelites are used to prepare combs, pens etc...

(i) The biochemical processes such as metabolism, cell-signaling etc... are essential for the normal functioning of our body.
(ii) These routine processes may be disturbed by any external factors such as microorganism, chemicals etc ..
(iii) Under such conditions medicines restore the normal functioning of the body.
(iv) These drug molecules interact with biomolecules such as proteins, lipids, etc .. that are responsible for different functions of the body.
(v) The drug interacts with these molecules and modify the normal biochemical reactions either by modifying the enzyme activity or by stimulating/suppressing certain receptors.

The intermediate compound formation theory :
A catalyst acts by providing a new path with low energy of activation. In homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the catalyst is regenerated.
Consider the reactions :
A+B➝AB; C is the catalyst .............(1)
A + C ➝ AC (intermediate) ..........(2)
AC + B ⟶ AB + C ...............(3)
Example 1:
The mechanism of Fridel crafts reaction is given below
\({ C }_{ 6 }{ H }_{ 6 }+{ { CH }_{ 3 }Cl\overset { anhydrous\\ { AlCl }_{ 3 } }{ \longrightarrow } }{ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 3 }+HCl\) 
The action of catalyst is explained as follows
CH₃Cl + AlCl₃ ⟶ [CH₃]⁺ [AlCl₄]^-
It is an intermediate.
\({ C }_{ 6 }{ H }_{ 6 }+\left[ { CH }_{ 3 }^{ + } \right] \left[ { AlCl }_{ 4 } \right] ^{ - }\longrightarrow { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 3 }+{ AlCl }_{ 3 }+Hcl\) 
Example 2:
\({ { 2KClO }_{ 3 }\overset {\\ { MnO }_{ 3 } }{ \longrightarrow } }{ 2KCl }+{ 3O }_{ 2 }\)
Thermal decomposition of KCIO₃ in the presence of MnO₂ proceeds as follows. Steps in the reaction
2KCIO₃ ⟶ 2KCI + 3O₂ Can be given as
2KClO₃ + 6MnO₂   →  6MnO₃ + 2KCl
It is an intermediate
6MnO₃ → 6MnO₂ + 3O₂
Example 3:
Formation of water due to the reaction of H₂ and O₂ in the presence of Cu can be given as
H₂ + 1/2O₂ → H₂O 
2Cu + \(\frac{1}{2}\)O₂   → Cu₂O
It is an intermediate.
Cu₂O + H₂  → H₂O + 2Cu
Advantages:
This theory describes
(a) The specificity of a catalyst and
(b) The increase in the rate of the reaction with increasc inthe concentration of a catalyst
Limitations:
(a) The intermediate compound theory fails to explain the action of catalytic poison and activators (promoters).
(b) This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Hoffmann's bromide reaction:
When Amides are treated with bromine in the presence of aqueous or ethanolic solution of KOH, primary amines with one carbon atom less than the parent amides are obtained.
\(\underset { \quad \quad \quad amide\\ R=Alkyl(or)Aryl }{ R-\overset { \underset { || }{ O } }{ C } -{ NH }_{ 2 } } \overset { { Br }_{ 2 }/KOH }{ \longrightarrow } \underset { Primary\quad amine }{ R-{ NH }_{ 2 }+{ K }_{ 2 }{ CO }_{ 3 } } +KBr+{ H }_{ 2 }O\)
(ii) Hoffmann's ammonolysis:
When Alkyl halides (or) benzylhalides are heated with alcoholic ammonia in a sealed tube, mixtures of 1°, 2° and 3° amines and quaternary ammonium salts are obtained.
\( { CH }_{ 3 }-Br\overset { \ddot { N } { H }_{ 3 } }{ \underset { \Delta }{ \longrightarrow } } \underset { { 1 }^{ 0 }-amine }{ { CH }_{ 3 }-\ddot { N } { H }_{ 2 } } \overset { { CH }_{ 3 }-Br }{ \longrightarrow } \underset { { 3 }^{ o }-amine }{ \left( { CH }_{ 3 } \right) _{ 2 }\ddot { N } H } \overset { { CH }_{ 3 }Br }{ \longrightarrow } \underset { 3^{ o }-amine }{ \left( { { CH }_{ 3 } } \right) _{ 3 }\ddot { N } } \overset { CH_{ 3 }Br }{ \longrightarrow } \underset { Quartenary \ ammonium\ bromide }{ \left( { CH }_{ 3 } \right) _{ 4 }\overset { + }{ N } { Br }^{ - } } \) 
This is a nucleophilic substitution, the halide ion of alkyl halide is substituted by the -NH₂ group. The product primary amine so formed can also has a tendency to act as a nucleophile and hence if excess alkyl halide is taken, further nucleophilic substitution takes place leading to the formation of quarternary ammonium salt. However, if the process is carried out with excess ammonia, primary amine is obtained as the major product. The order of reactivity of alkylhalides with amines
RI > RBr > RCl
(iii) Gabriel phthalimide synthesis:
Gabriel synthesis is used for the preparation of Aliphatic primary amines. Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine. Aniline cannot be prepared by this method because the arylhalides do not undergo nucleophilic substitution with the anion formed by phthalimide.

(iv) Schotten - Baumann reaction :
Aniline reacts with benzoylchloride (C₆H₅COCI) in the presence of NaOH to give N - phenyl benzamide. This reaction is known as Schotten - Baumann reaction. The acylation and benzoylation are nucleophilic substitutions.
\(\underset { Aniline }{ { C }_{ 6 }{ H }_{ 5 }-{ NH }_{ 2 } } +\underset { Benzoyl\quad chloride }{ { C }_{ 6 }{ H }_{ 5 }-\overset { \underset { || }{ O } }{ C } -Cl } \overset { Pyridine }{ \longrightarrow } \underset { N-phenyl\quad benzamide }{ { C }_{ 6 }{ H }_{ 5 }-NH-\overset { \underset { || }{ O } }{ C } -{ C }_{ 6 }{ H }_{ 5 } } +HCl\)
(v) Carbylamine reaction :
Aliphatic (or) aromatic primary amines react with: chloroform and alcoholic KOH to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test used to identify the primary amines.
\(\underset { Ethylamine }{ { C }_{ 2 }{ H }_{ 5 }-{ NH }_{ 2 } } +\underset { Chloroform }{ { CHCl }_{ 3 }+3KOH } \longrightarrow \underset { Ethylisocyanide }{ { C }_{ 2 }{ H }_{ 5 }-NC } +3KCl+3{ H }_{ 2 }O\)
(vi) Mustard oil reaction:
(a) When primary amines are treated with carbon disulphide (CS₂), N - alkyldithio carbonic acid is formed which on subsequent treatment with HgCI₂, gives an alkyl isothiocyanate.


(b) When aniline is treated with carbon disulphide, or heated together, S-diphenylthio urea is formed, which on boiling with strong HCI, phenyl isothiocyanate (phenyl mustard oil), is formed.

These reactions are known as Hofmann - Mustard oil reaction. This test is used to identify the primary amines.
(vii) Coupling reactions (or) p-hydroxyazobenzene, p-aminoazobenzene, 2-phenyl azo-4-methyl phenol:
Benzene diazonium chloride reacts with electron rich aromatic compounds like phenol, aniline to form brightly coloured azo compounds. Coupling generally occurs at the para position. If para position is occupied then coupling occurs at the ortho position. Coupling tendency is enhanced if an electron donating group is present at the para - position to -N₂CI- group. This is an electrophilic substitution.

Aryl fluorides and iodides cannot be prepared by direct halogenation and the cyano group cannot be introduced by nucleophilic substitution of chlorine in chlorobenzene. For introducing such a halide group. cyano group -OH, NO₂, etc.. benzenediazonium chloride is a very good intermediate Diazo compounds obtained from the coupling reactions of diazonium salts are coloured and are used as dyes.
(viii) Diazotisation :
Aniline reacts with nitrous acid at low temperature (273 - 278 K) to give benzene. Diazonium chloride which is stable for a short time and slowly decomposes even at low temperatures.This reaction is known as diazotization

(ix) Gomberg reaction :
Benzene diazonium chloride reacts with benzene in the presence of sodium hydroxide to give biphenyl. This reaction in known as the Gomberg reaction

A drug is a substance that is used to modify or explore physiological systems or pathological states for the benefit of the recipient. It is used for the purpose of diagnosis, prevention, cure/relief of a disease.
a) Classification of drugs:
Drugs are classified based on their properties such as chemical structure, pharmacological effect, target system, site of action etc.
b) Classification based on the chemical structure:
In this classification, drugs with a common chemical skeleton are classified into a single group. For example, ampicillin, amoxicillin, methicillin etc.. all have similar structure and are classified into a single group called penicillin. Similarly, we have other group of drugs such as opiates, steroids, catecholamines etc. Compounds having similar chemical structure are expected to have similar chemical properties. However, their biological actions are not always similar. For example, all drugs belonging to penicillin group have same biological action, while groups such as barbiturates, steroids etc.. have different biological action.
Penicillins

Classification based on Pharmacological effect:
In this classification, the drugs are grouped based on their biological effect that they produce on the recipient. For example, the medicines that have the ability to kill the pathogenic bacteria are grouped as antibiotics. This kind of grouping will provide the full range of drugs that can be used for a particular condition (disease). The physician has to carefully choose a suitable medicine from the available drugs based on the clinical condition of the recipient.
Examples:
Antibiotic drugs: amoxicillin, ampicillin, cefixime, cefpodoxime, erythromycin, tetracycline etc.. Antihypertensive drugs: propranolol, atenolol, metoprolol succinate, amlodipine etc...
Classification based on the target system (drug action):
In this classification, the drugs are grouped based on the biological system/process, that they target in the recipient. This classification is more specific than the pharmacological classification. For example, the antibiotics streptomycin and erythromycin inhibit the protein synthesis (target process) in bacteria and are classified in a same group. However, their mode of action is different. Streptomycin inhibits the initiation of protein synthesis, while erythromycin prevents the incorporation of new amino acids to the protein.
Classification based on the site of action (molecular target):
The drug molecule interacts with biomolecules such as enzymes, receptors etc, which are referred as drug targets. We can classify the drug based on the drug target with which it binds. This classification is highly specific compared to the others. These compounds often have a common mechanism of action, as the target is the same.

1. The crude metal is refined by electrolysis. It is carried out in an electrolytic cell
Anode : Impure metal to be refined with dilute acid.
Cathode : Thin strips of pure metal
Electrolyte : Aqueous solution of the salts of the metal with dilute acid.
2. The metal dissolves from the anode, pass into the solution.
3. At the same amount of metal ions from the solution will be deposited at the cathode.
4. During electrolysis, the less electropositive impurities in the anode, settle down at the bottom and are removed as anode mud.
Example: Electrolytic refining of silver.
Cathode: Pure silver
Anode: lmpure silver rods
Electrolyte: Acidified aqueous solution of silver nitrate
5. When a current is passed through the electrodes the following reactions will take place
(a) Reaction at anode: \({ Ag }_{ (s) }\longrightarrow { Ag }^{ + }_{ (aq) }+{ 1e }^{ - }\)
(b) Reaction at cathode: \({ Ag }^{ + }_{ (aq) }+{ 1e }^{ - }\longrightarrow { Ag }_{ (s) }\)
6. During electrolysis, at anode silver loses electrons and form silver ions and the silver ions migrate towards the cathode and get discharged and deposited on the cathode.
7. Copper, Zinc etc can also be refined by this process.

Given: [Cu²⁺] = 0.1M; E⁰_Cu²⁺|Cu = +0.34V
Cell reaction is: \(Cu^{2+}_{(aq)}+2e^{-}\rightarrow Cu_{(s)}\)
\(E_{cell}=E^{0}-\frac{0.0591}{n}\log\frac{[Cu]}{[Cu^{2+}]}\)
\(= 0.34\frac{0.0591}{2}\log\frac{1}{0.1}\)
E_cell = 0.34 - 0.0296 = +0.31V

Anodic oxidation: 
The electrode at which the oxidation occurs is called the anode. In  Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to Zn²⁺ ions and the electrons. The Zn²⁺ ions enters the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip. Electrons are liberated at zinc electrode and hence it is negative (-ve).
\({ Zn }_{ (s) }\longrightarrow { Zn }^{ 2+ }{ _{ (aq) } }+ 2{ e }^{ - } \) (loss of electron-oxidation) 
Cathodic reduction:
The electrons flow through the circuit from zinc to copper, where the Cu²⁺ ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ve).
\({ Cu }^{ 2+ }_{ (aq) }+{ 2e }^{ - }\longrightarrow { Cu }_{ (s) } \)(gain of electron-reduction) 

\(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightleftharpoons 3 \mathrm{Ca}^{2+}_{(aq)}+2 \mathrm{PO}_{4_{(aq)}}^{3-}\\\)
(s)                           (3s)             (2s)
\(K_{sp}=[Ca^{2+}]^{3}[PO_{4}^{3-}]^{2}\)
\(K_{sp}=(3s)^{3}(2s)^{2}\)
\(K_{sp}=27s^{3}.4s^{2}\)
\(K_{sp}=108s^{5}\)
(or)
\(K_{s p} =m^{m} \cdot n^{n} \cdot(s)^{m+n} \)
\(K_{\text {sp }} =3^{3} \cdot 2^{2} \cdot(s)^{3+2} \)
\(K_{s p} =27 \times 4 \times(s)^{5} \)
\(=108(s)^{5}=108 s^{5}\)

(i) They are hard and show electrical land thermal conductivity.
(ii) They have high melting points higher than those of pure metals.
(iii) Transition metal hydrides are used as powerful reducing agents.
(iv) Metallic carbides are chemically inert

(i) When the concentration of the reactant is increased by 'n' times, the rate of reaction is also increased by n times. That is, if the concentration of the reactant is doubled, the rate is doubled.
(ii) The unit of rate constant of a first order reaction is sec^-1 or time^-1.
\({ k }_{ 1 }=\frac { rate }{ (a-x) } =\frac { { mol.lit }^{ -1 }{ sec }^{ -1 } }{ { mol.lit }^{ -1 } } \)
(iii) The time required to complete a definite fraction of reaction is independent of the initial concentration, of the reactant if t_1/u is the time of one 'u' th fraction of reaction to take place then from equation.
\({ k }_{ 1 }=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
\(x=\frac { a }{ u } and\quad { t }_{ \frac { 1 }{ u } }=\frac { 2.303 }{ { k }_{ 1 } } \log { \frac { a }{ a-\frac { a }{ u } } } ;\)
\({ t }_{ \frac { 1 }{ u } }=\frac { 2.303 }{ { k }_{ 1 } } \log { \frac { u }{ (u-1) } } \)
since k₁ = rate constant, t_1/u is independent of initial concentration 'a'.

(i) PCI₅+ C₂H₅OH \(\longrightarrow \) C₂H₅CI + POCl₃ + HCI
(ii) P₄+ 10SO₂Cl₂ \(\longrightarrow \) 4PCI₅ + 10SO₂
(iii) POCl₃ + 3H₂O \(\longrightarrow \) H₃PO₄ + 3HCI
(iv) PH₃ + HBr\(\longrightarrow \) PH₄Br
(v) PCl₃+ 3H₂O\(\longrightarrow \) H₃PO₃ + 3 HCI

(i) Fluorine is the most reactive element among halogens due to minimum value of F- F bond dissociation energy.
(ii) It can form two types of salts with metals NaF and NaHF₂
(iii) AgF is soluble in water but other AgX are insoluble.
(iv) HF attacks glass while others do not.
(v) Fluorine, does not form any polyhalides (absence of d - orbitals)
(vi) Fluorine exhibit only negative oxidation state (highly electronegative) while other halogens have both +ve and -ve oxidation state.

O₃ oxidises potassium iodide to iodine. This reaction is quantitative and can be used for estimation of ozone 
\({ O }_{ 3 }+2KI+{ H }_{ 2 }O\longrightarrow 2KOH+{ O }_{ 2 }+{ I }_{ 2 }\)

(i) CaC₂
2 + 2x = 0
2x = -2
x = -1
(ii) H₂CO₃
2 + x + (-6) = 0
x = +4
(iii) HCN
1 + x + (-3) = 0
x = +3 - 1 = +2
(iv) CO
x + (-2) = 0
x= +2

(i) Boron trifluoride is obtained by the treatment of calcium fluoride with boron trioxide in presence of conc. sulphuric acid.
B₂O₃ + 3CaF₂ + 3H₂SO₄ \(\overset { \triangle }{ \longrightarrow } \) 2BH₃ + 3CaSO₄ + 3H₂O
(ii) It can also be obtained by treating boron trioxide with carbon and fluorine.
B₂O₃ + 3C + 3F₂ ⟶ 2BF₃ + 3CO
(iii) In the laboratory pure BF₃ is prepared by the thermal decomposition of benzene diazonium tetrafluoro borate.
PhN₂BF₄ \(\overset { \triangle }{ \longrightarrow } \) BF₃ + PhF + N₂

(i) Gold, one of the expensive and precious metals. It is used for coinage, and has been used as standard for monetary systems in some countries.
(ii) It is used extensively in jewellery in its alloy form with copper.
(iii) It is also used in electroplating to cover other metals with a thin layer of gold which are used in watches, artificial limb joints, cheap jewellery, dental fillings and electrical connectors.
(iv) Gold nanoparticles are also used for increasing the efficiency of solar cells and also used an catalysts.

(i) Metallic oxides such as Cr₂O₃ can be reduced by an aluminothermic process.
(ii) In this process, the metal oxide is mixed with aluminum powder and placed in a fire clay crucible.
(iii)To initiate the reduction process, an ignition mixture (usually magnesium and barium peroxide is used 
\({ BaO }_{ 2 }+Mg\longrightarrow Bao+MgO\)
(iv) During the above reaction a large amount of heat is evolved (temperature up to 2400°C, is generated and the reaction enthalpy is: 852 kJ mol^-1 which facilitates the reduction of Cr₂O₃ by aluminium power.
\({ Cr }_{ 2 }{ O }_{ 3 }+2Al\overset { \Delta }{ \longrightarrow } 2Cr+{ Al }_{ 2 }{ O }_{ 3 }\)

(i) Roasting is the method, usually applied for the conversion of sulphide ores into their oxides.
(ii) In this method, the concentrated ore is oxidised by heating it with excess of oxygen in a suitable furnace below the melting point of the metal.
(iii) \(2PbS+{ 3O }_{ 2 }\overset { \Delta }{ \longrightarrow } 2PbO+{ 2SO }_{ 2 }\uparrow \)
\(2ZnS+{ 3O }_{ 2 }\overset { \Delta }{ \longrightarrow } 2ZnO+2{ SO }_{ 2 }\uparrow \)
\(2Cu_{ 2 }S+{ 3O }_{ 2 }\overset { \Delta }{ \longrightarrow } 2Cu_2O + 2SO_2\uparrow \)
(iii) Roasting also removes impurities such as arsenic, sulphur, phosphorous by converting them into their volatile oxides
(v) Ex.\(4As+{ { 3O }_{ 2 } }\longrightarrow { 2As }_{ 2 }{ O }_{ 3 }\uparrow \)
\( \mathrm{S}_{8}+8 \mathrm{O}_{2} \longrightarrow 8 \mathrm{SO}_{2} \uparrow \)
\(\mathrm{P}_{4}+5 \mathrm{O}_{2} \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10} \uparrow\)

If we perform a qualitative analysis to identify the constituent ions present in both the compounds, Mohr's salt answers the presence of Fe²⁺, NH₄^+, and SO₄^2- ions, whereas the potassium ferrithiocyanate will not answer Fe³⁺ and SCN ions. We can infer that the double salts loose their identity and dissociate into their constituent simple ions in solutions, whereas the complex ion in coordination compound, does not loose its identity and never dissociate to give simple ions.

For a first order reaction
\(k=\frac { 2.303 }{ t } \log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(k=\frac { 2.303 }{ t } \log\frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ 1 } } \)
V_∞= 58.3 ml.

As the rate constants are constant through out it is a first order reaction. 

\(\mathrm{k}=0.693 / \mathrm{t}_{1 / 2}\)
\(\mathrm{k}=\frac{0.693}{8.0}=0.08 \mathrm{~min}^{-1}\)
For the first order reaction:
\(t =\frac{2.303}{k} \log \frac{\left[A_{0}\right]}{[A]}  \)
\(t =\frac{2.303}{0.087} \log \left(\frac{100}{1}\right)=26.47 \log 10^{2}  \)
\(t =2 \times 26.47 \log 10  \)
\(t =52.94 \mathrm{~min}\)

(i) Reaction Rate = \(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 }\) ...(1)
When [L] = [4L]
Rate = \(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ 4L \right] }^{ 3/2 }\)
The Reaction Rate = \(8(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 })\) ...(2)
Comparing (1) and (2) rate is increased by 8 times
(ii)  [A] = [2A] and [B] = [2B]
Reaction Rate = \(k{ \left[ 2A \right] }^{ 2 }\left[ 2B \right] { \left[ L \right] }^{ 3/2 }\)
Reaction Rate = \(\\ 8(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 })\)  ...(3)
Comparing (1) and (3) rate is increased by 8 times
(iii) \(\left[ A \right] =\left[ \frac { A }{ 2 } \right] \)
Reaction Rate = \(k{ \left[ \frac { A }{ 2 } \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 }\)
Reaction Rate = \(\frac { 1 }{ 4 } \left( k\left[ { A }^{ 2 } \right] { \left[ B \right] }{ \left[ L \right] }^{ 3/2 } \right) \) ...(4)
Comparing (1) and (4); rate is reduced to (1/4) times.
(iv) \(\left[ A \right] =\left[ \frac { 1 }{ 3 }A \right] and\left[ L \right] =\left[ 4L \right] \)
Rate = \(k{ \left[ \frac { 1 }{ 3 }A \right] }^{ 2 }{ \left[ B \right] }{ \left[ 4L \right] }^{ 3/2 }\)
Rate = \(\left( \frac { 8 }{ 9 } \right) \left( k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 } \right)\) ...(5)
Comparing (1) and (5); rate is reduced to \(\frac { 8 }{ 9 } \) times.

(i) The half life of a reaction is defined as the time required for the reactant concentration to reach one half its initial value.
(ii)   The rate constant for a first order reaction is given by
\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(at\quad t={ t }_{ \frac { 1 }{ 2 } };\left[ A \right] =\frac { \left[ { A }_{ 0 } \right] }{ 2 } \)
\(k=\frac { 2.303 }{ t_{ 1/2 } } log\frac { \left[ { A }_{ 0 } \right] }{ \frac { \left[ { A }_{ 0 } \right] }{ 2 } } \)
\(k=\frac { 2.303 }{ { t }_{ 1/2 } } log2\)
\(k=\frac { 2.303\times 0.3010 }{ { t }_{ 1/2 } } =\frac { 0.6932 }{ { t }_{ 1/2 } } \)
\({ t }_{ 1/2 }=\frac { 0.6932 }{ k } \)
This equation has no concentration term So, the half life of a first order reaction is independent of initial concentration.

Given: NaCl is doped with 10^-2 mole % of SrCl₂
(i.e.) 100 moles of NaCl doped with 10^-2 moles of SrCl₂
\(\therefore \) 1 mole of NaCl is doped with 10^-4 moles of SrCl₂
1 Sr²⁺  ion creates 1 cation vacancy
The number of cation vacancies created by 10^-4 mole SrCl₂ = 10^-4 \(\times\) 6.023 \(\times\) 10²³
= 6.023 \(\times\) 10¹⁹ vacancies

There are seven types of unit cell, Cubic, tetragonal, orthorhombic, hexagonal, monoclinic, triclinic and rhombohedral. They differ in the arrangement of their crystallographic axes and angles.
i) Cubic: a = b = c; α = β = ૪ = 90^o.

ii) Tetragonal: a = b ≠ c; α = β = ૪ = 90°.

iii) Orthorhombic:  a  ≠ b  ≠ c; α = β = ૪ = 90°.

iv) Hexagonal: a = b ≠ c; α = β = 90^o, ૪ = 120^o. 

v) Monoclinic: a ≠ b ≠ c; α = ૪ = 90^o, β ≠ 90^o, 

vi) Triclinic: a ≠ b ≠ c; α  ≠ β ≠ ૪ ≠ 90^o.

vii) Rhombohedral:  a = b = c; α = β = ૪ ≠ 90^o. 

2KClO₃ + 6MnO₂   →  6MnO₃ + 2KCl
It is an intermediate
6MnO₃ → 6MnO₂ + 3O₂

Addition of electrolytes:
A negative ion causes the precipitation of positively charged sol and vice versa. When the valency of ion is high, the precipitation power is increased.
For example, the precipitation power of some cations and anions varies in the following order
\(\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^{+} \text {, Similarly }\left[\mathrm{Fe}\left(\mathrm{CN}_{6}\right)\right]^{3-}>\mathrm{SO}_{4}{ }^{2-}>\mathrm{Cl}^{-}\)
The precipitation power of electrolyte is determined by finding the minimum concentration (millimoles/lit) required to cause precipitation of a sol in 2 hours. This value is called flocculation value. The smaller the flocculation value greater will be precipitation.

a) Primary and secondary nitroalkanes, having α-H, also show an equilibrium mixture of two tautomers namely nitro - and aci - form

b) Difference:

An alcohol:

An alkylhalide:

An alkene:

\(\mathrm{Hg}_{2} \mathrm{Cl}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Hg}_{2}^{2+} \text { (aq) }+2 \mathrm{Cl^-}_{(\mathrm{aq})}\\ s \quad \quad \quad \quad \quad \quad s \quad \quad \quad \quad \quad 2s\)
\(\mathrm{K}_{\mathrm{sp}} =\left[\mathrm{Hg}_{2}^{2+}\right]{\left[\mathrm{Cl}^{-}\right]^{2}}  \)
\(=(\mathrm{s})(2 \mathrm{~s})^{2} \)
\(\mathrm{~K}_{\mathrm{sp}} =4 \mathrm{~s}^{3}\)

Aspirin is o-acetyl salicylic acid

Conjugate Pairs:
\((a) \mathrm{HS}_{\text {(aq) }}^{-} \& \mathrm{H}_{2} \mathrm{~S}_{\text {(aq) }} \) \((b) \mathrm{HF}_{\text {(aq) }} \& \mathrm{~F}_{\text {(aq) }}^{-} \)
\((a) \mathrm{HPO}_{4}^{2-} \& \mathrm{PO}_{4}^{3-} \) \((b) \mathrm{SO}_{3}^{2-} \& \mathrm{HSO}_{3}^{-} \)
\((a) \mathrm{NH}_{4}^{+} \& \mathrm{NH}_{3} \) \((b) \mathrm{CO}_{3}^{2-} \& \mathrm{HCO}_{3}^{-}\)

(i) I Order reaction
(ii) S^-1
(iii) \({ t }_{ \frac { 1 }{ 2 } }\) = 0.693/k

No. of atoms of X at the corners = 8 - 1 = 7
(since 1 is missing)
Contribution of X towards unit cell
\(=7\times \frac { 1 }{ 8 } =\frac { 7 }{ 8 } \)
No. of atoms of Y at the face centres = 6
Contribution of Y towards unit cell
= 6 x \(\frac{1}{2}\) = 3
Ratio of X : Y = \(\frac{7}{8}\) : 3 = 7:24
Formula of the compound is X₇ Y₂₄.

(i) Zeolites have a three dimensional crystalline structure looks like a honeycomb consisting of a network of interconnected tunnels and cages.
(ii) Water molecules moves freely in and out of these pores hence they are highly porous.

Transition elements have strong tendency to form:
complexes because of two reasons. They are
(i) Small size and high positive charge density.
(ii) Presence of cant (n-1) d orbitals which are of appropriate energy to accept lone pair and unshared pair of electrons from the ligands for bonding with them.

Sodium cyanide and sodium carbonate.

SiO₂ is used in the metallurgy of copper to remove FeO as FeSiO₃ (slag) (i.e) acidic flux is used to remove basic impurities.
\(FeO+{ SiO }_{ 2 }\longrightarrow { FeSiO }_{ 3 }\)

Chromium (24), the electronic configuration is 3d⁵4s¹. It ready to lose its outer most electron (4s¹) to get exactly half-filled stable electronic configuration. The electronic configuration of Zinc is 3d¹⁰4s².ie., It has completely filled stable configuration. From this configuration, the removal of 1e^- from 4s orbital is very difficult & it required more ionisation enthalpy. Due to this reason Zn has higher first Ionisation enthalpy (1.E₁) than that of chromium.

Oxidation number of F = -1
Oxidation number of H = +1
Oxidation number of O in HOF =x
(+1) + x + (-1) = 0
x = 0
Oxidation number of O in HOF = 0

B) if both assertion and reason are true but reason is not the correct explanation of assertion.