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

pH = -log₁₀ [H⁺]
∴  [H⁺] = 10^-pH
Let the volume be x ml.
V₁M₁ +V₂M₂ +V₃M₃ = VM
∴   x ml of 10^-1 M+ x ml of 10^-2 M + x ml of 10^-3 M
= 3 x ml of [H⁺]
\(∴  [H^+] = \frac{ x[0.1 + 0.01 + 0.001]}{3x}\)
\(= \frac{ [0.1 + 0.01 + 0.001]}{3}\)
\(= \frac{ [0.111]}{3}\)
= 0.037
= 3.7 x 10^-2

If a is the length of the side, then the length of the leading diagonal passing through the body centered atom is \(\sqrt{3a} \)
Required distance =  \(\left( \cfrac { \sqrt { 3 } }{ 2 } \right) a\)

4% solution of ethylnitrite in alcohol is known as sweet spirit of nitre and in used as diuretic.

Pure acetic acid is called glacial acetic acid. Because it forms ice like crystal when cooled. When aqueous acetic acid is cooled at 289.5 K, acetic acid solidifies and forms ice like crystals, where as water remains in liquid state and removed by filtration. This process is repeated to obtain glacial acetic acid.

Acetonitrile (CH₃CN) is hydrolysed with aqueous acid or alkali to get acetic acid.
\(\\ \\ \underset { Methylcyanide\\ \quad \quad (or)\\ Acetonitrile }{ { CH }_{ 3 }-C\equiv N+{ H }_{ 2 }O } \longrightarrow \underset { Acetamide }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ NH }_{ 2 } } \overset { { H }_{ 2 }O }{ \longrightarrow } \underset { Acetic\ acid }{ { CH }_{ 3 }COOH+{ NH }_{ 3 } } \)

Diisobutyl aluminium hydride (DIBAL -H) selectively reduces the alkyl cyanides to form imines which on hydrolysis gives aldehydes.
Example :
\(\underset { hex-4-ennitrile }{ { CH }_{ 3 }-CH=CH-{ CH }_{ 2 }-{ CH }_{ 2 }-CN } \overset { i)AIH\left( t-bu \right) _{ 2 } }{ \underset { ii){ H }_{ 2 }o }{ \longrightarrow } } \underset { hex-4-enal }{ { CH }_{ 3 }-CH=CH-{ CH }_{ 2 }-CHO } \) 

(i) 
(ii) 

Invertase or sucrose :
\(\underset { Sucrose }{ { C }_{ 12 }{ H }_{ 22 }{ O }_{ 11 } } +{ H }_{ 2 }O\overset { Invertase }{ \underset { orSucrose }{ \longrightarrow } } \underset { Glucose }{ { C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 } } +\underset { Frutose }{ { C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 } } \)

Starch consists of two fraction
(i) amylose and
(ii) amylopectin

(i) Disaccharides linked through C₁ of the first to C₄ or C₆ of the second component are called reducing sugar. Eg: Lactose.
(ii) Disaccharides linked through the glycosidic carbon atoms of each component C₁ of glucose to C₂ of fructose are nonreducing disaccharides. Eg: Sucrose

The function of tRNA is to carry amino acids to the sites of protein synthesis on ribosomes.

In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pair of bases. Cytosine forms hydrogen bond with guanine while adenine forms hydrogen bonds with thymine. As a result the two strands are complementary to each other.

Nucleotides are monomers of nucleic acid, They consist of heterocyclic base, pentose sugar and phosphoric acid residue.

Ethers having same molecular formula, but alkyl groups attached to the oxygen atom are different. This phenomenon is known as metamerism.
Metamers of 2- methoxy propane
\(\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{O}-\mathrm{C}_{2} \mathrm{H}_{5}-\) diethyl ether 
\(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) methyl n-propyl ether

(a) B(OH)₃ + NH₃\(\overset { \Delta }{ \longrightarrow } \) BN + 3H₂O
                                (Boron nitride)
(b) Na₂B₄O₇ + H₂SO₄ + 5H₂O \(\longrightarrow \) 4H₃BO₃ + Na₂SO₄
                                                           (Boric acid)
(c) B₂H₆ + 2NaOH + 2H₂O \(\longrightarrow \)2NaBO₂ + 6H₂
                                              (Sodium metaborate)
(d) B₂H₆ + 6CH₃OH  \(\longrightarrow \)2B(OCH₃)₃ + 6H₂O
                                              (Trimethyl borate)
(e) 4BF₃ + 3H₂O \(\longrightarrow \) H₃BO₃ + 3H⁺ + 3[BF₄]^-
                                      (Boric acid)
(f) HCOOH + H₂SO₄ \(\longrightarrow \)CO + H₂SO₄. H₂O
                                        (Carbon monoxide)   
(g) 2SiCl₄ + NH₃ \(\overset { 330K }{ \underset { ether }{ \longrightarrow } } \)Cl₃Si- NH - SiCl₃ + 2HCl
                                        (Chlorosilazane)
(h) SiCl₄ + 4C₂H₅OH\(\longrightarrow \)Si(OC₂H₅)₄ + 4HCI
                                        (Tetraethoxysilane)
(i) 2B + 6NaOH\(\longrightarrow \) 2Na₃BO₃ + 3H₂
(j) H₂B₄O₇ \(\xrightarrow[]{Redhot}\) 2B₂O₃ + H₂O

(i) Icosogens → B, Al, Ga, In, Tl
(ii) Tetragens → C, Si, Ge, Sn, Pb
(iii) Pnictogens → N, P, As, Sb, Bi
(iv) Chalcogens → O, S, Se, Te, Po

In this method, the metal is treated with a suitable reagent which can form a volatile compound with the metal.
Then the volatile compound is decomposed to give the pure metal.

(i) An organic compound (A) is identified as C₆H₅CHO benzaldehyde. Benzaldehyde reduces Tollen'sreagent and also undergoes.
Cannizaro reaction:
\(\\ \underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO } +{ Ag }_{ 2 }O\longrightarrow 2Ag+{ C }_{ 6 }{ H }_{ 5 }COOH\) 
Cannizaro reaction:
\({ C }_{ 6 }{ H }_{ 5 }CHO+{ C }_{ 6 }{ H }_{ 5 }CHO\overset { NaOH }{ \longrightarrow } { C }_{ 6 }{ H }_{ 5 }{ OH }OH+\underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }COOH } \) 
(ii) Benzaldehyde on oxidation gives Benzoic acid C₆H₅COOH and it is (B).
\({ C }_{ 6 }{ H }_{ 5 }CHO\overset { (O) }{ \longrightarrow } \underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }COOH } \) 
(iii) Calcium salt of benzoic acid (calcium benzoate) on dry distillation gives benzophenone C₆H_sCOC₆H_s as (C).

(i) Hydrolysis: Acid anhydride are slowly hydrolysed, by water to form corresponding carboxylic acids.

(ii) Reaction with alcohol:

\(\underset { Acetaldehyde }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -H+4(H) } \overset { { NH }_{ 2 }{ NH }_{ 2 } }{ \underset { { C }_{ 2 }{ H }_{ 5 }ONa }{ \longrightarrow } } \underset { Ethane }{ { CH }_{ 3 }-{ CH }_{ 3 }+{ H }_{ 2 }O+{ N }_{ 2 } } \)
(i) Clemmensen reduction: Aldehydes and Ketones when heated with zinc amalgam and concentrated hydrochloric acid gives hydrocarbons. 
Example:
\(\underset { Acetald4ehyde }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ H }^{ + }4(H) } \overset { Zn^{ - }Hg }{ \underset { Conc.HCl }{ \longrightarrow } } \underset {Ethane}{{ CH }_{ 3 }-{ CH }_{ 3 }\pm { H }_{ 2 }O}\) 
\(\underset { Acetone }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }+4\left( H \right) } \overset { { Zn }^{ - }Hg }{ \underset { Conc.HCl }{ \longrightarrow } } { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 3 }+{ H }_{ 2 }O\)
(ii) Wolff-Kishner Reduction: Aldehydes and Ketones when heated with hydrazine (NH₂NH₂) and sodium ethoxide, hydrocarbons are formed Hydrazine acts as a reducing agent and sodium ethoxide as a catalyst.
Example :
\(\underset { Acetaldehyde }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -H+4(H) } \overset { { NH }_{ 2 }{ NH }_{ 2 } }{ \underset { { C }_{ 2 }{ H }_{ 5 }ONa }{ \longrightarrow } } \underset { Ethane }{ { CH }_{ 3 }-{ CH }_{ 3 }+{ H }_{ 2 }O+{ N }_{ 2 } } \) 
\(\underset { Acetone }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }+4(H) } \overset { { NH }_{ 2 }{ NH }_{ 2 } }{ \underset { { C }_{ 2 }{ H }_{ 5 }ONa }{ \longrightarrow } } { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 3 }+{ H }_{ 2 }O+{ N }_{ 2 }\)
Aldehyde (or) ketones is first converted to its hydrazone which on heating with strong base gives hydrocarbons.
(iii) Reduction to pinacols: Ketones, on reduction with magnesium amalgam and water, are reduced to symmetrical diols known as pinacol.

Importance of carbohydrates:
(i) Carbohydrates, widely distributed in plants and animals, acts mainly as energy sources and structural polymers.
(ii) Carbohydrate is stored in the body as glycogen and in plant as starch.
(iii) Carbohydrates such as cellulose which is the primary components of plant cell wall, is used to make paper, furniture (wood) and cloths (cotton)
(iv) Simple sugar glucose serves as an instant source of energy.
(v) Ribose sugars are one of the components of nucleic acids.
(vi) Modified carbohydrates such as hyaluronate (glycosaminoglycans) act as shock absorber and lubricant.

(i) Free radical polymerisation.
(ii) Styrene polymerises to polystyrene when it is heated to ionic with a peroxide initiator. The mechanism involves the following steps.

(iii) Propagation step

The stabilized radical attacks another monomer molecule to give an elongated radical.

Chain growth will continue with the successive addition of several thousands of monomer units.
(iv) The above chain reaction can be stopped by stopping the supply of monomer or by coupling of two chains or reaction with an impurity such as oxygen.

(i) Many drugs exert their physiological effects by binding to a specific molecule called a receptor whose role is to trigger a response in a cell.
(ii) The chemical messengers, bind to the active site of these receptors. This brings about the transfer of message into the cell.
(iii) The chemical messengers bind to the active site of these receptors.
(iv) This brings about the transfer of message into the cell
(v) These receptors show high selectivity.
(vi) If we want to block a message, a drug that binds to the receptor site should inhibit its natural function.
(vii) Such drugs are called antagonists.
(viii) In contrast, there are drugs which mimic the natural messenger by switching on the receptor. These type of drugs are called agonists.

(i) Nitroethane exhibits tautomerism of nitro form and aci form. Aciform of nitroethane contains (α - Hydrogen) replaceable hydrogen atom. Hence it dissolves in sodium hydroxide solution forming salt like compounds.

(ii) CH₃CH₂NO₂ has two a-hydrogen and that reacts with HNO₂
(iii) 2-methyl - 2 - nitro propane neither soluble in alkali nor reacts with nitrous acid. Because it contains a tertiary nitro group where aciform is not possible and . it does not contain α - H atoms. So it has neither of the properties.

(i) Compound (A) is carbonyl compound, it is acetone
(ii) (A) on reduction with LiAlH₄ gives (B) it gives blue colour in Victor Meyer'stest.
\({ CH }_{ 3 }-\underset { \overset { || }{ \underset { (A) }{ O } } }{ C } -{ CH }_{ 3 }\overset { { LiAIH }_{ 4 } }{ \underset { \left[ H \right] }{ \longrightarrow } } { CH }_{ 3 }-{ CH }_{ 3 }-\underset { \overset { | }{ \underset { (B) }{ OH } } }{ CH } -{ CH }_{ 3 }\) 
(iii) (B) reacts with SOCl₂to give (C).

(iv) (C) on treatment with alcoholic KOH, forms (D) by elimination reaction.
\({ CH }_{ 3 }-\underset { \overset { | }{ \underset { (C) }{ Cl } } }{ C } H-{ CH }_{ 3 }\overset { alc.KOH }{ \longrightarrow } \underset { (D) }{ { CH }_{ 3 }CH={ CH }_{ 2 }+HCl } \) 

(i) For a chemical reaction, catalyst is needed in very small quantity.
(ii) There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.
(iii) A catalyst itself cannot initiate a reaction. 
(iv) A solid catalyst will be more effective if it is taken in a finely divided form.
(v) A catalyst are specific in nature.
(vi) In an equilibrium reaction, presence of catalyst reduces the time for attainment of equilibrium and hence it does not affect the position of equilibrium and the value of equilibrium constant.
(vii) A catalyst is highly effective at a particular temperature called as optimum temperature.
(viii) Presence of a catalyst generally does not change the nature of products

Given:
E₁ = 0.5 V
Cr³⁺+ 3e^- ➝ Cr_(s).......(1)
E₂ = 0.41 V
Cr³⁺+ e^- ➝ Cr²⁺  .....(2)
The required reaction is,
Cr²⁺+ 2 e^- ➝ Cr_(s)
Then,
Formula:
\({ E }_{ 3 }=\frac { 3{ E }_{ 1 }+{ E }_{ 2 } }{ 2 } \)
Solution:
= \(\frac { 3(0.5)+(0.41) }{ 2 } =\frac { 1.5+0.41 }{ 2 } \)
= 0.955 V
E₃ = 0.955 V

Degree if dissociation \(\alpha =\sqrt { \frac { K_{ a } }{ C } } ,C\alpha =\sqrt { { K }_{ a }.C } \)
∴ \(pOH=\log\frac { 1 }{ \sqrt { { K }_{ b }.C } } =\log\frac { { 10 }^{ 3 } }{ \sqrt { 1.75\times { 10 }^{ -5 }\times .1 } } \)
\(=\log\frac { 1 }{ \sqrt { 1.75\times { 10 }^{ -6 } } } =\log\frac { { 10 }^{ 3 } }{ \sqrt { 1.75 } } \)
\(=3-\frac { 1 }{ 2 } \log1.75\)
\(=3-\frac { 1 }{ 2 } \times 0.2430=2.8785\)
pH = 14 - pOH
∴ pH = 14 - 2.8785 = 11.1215
Alternating,
\(\left[ { H }^{ - } \right] =\sqrt { { K }_{ b }\times C } \)
\(=\sqrt { 1.75\times { 10 }^{ -5 }\times 0.1 } =\sqrt { 1.75\times { { 10 }^{ -6 } } } \)
= 1.322 x 10^-3
∴ pOH = - log [H⁺] = - log (1.323 x 10^-3)
3 - 0.1216 = 2.8784
∴ pH = 14 - pOH = 14 - 2.8784 = 11.1216

i) In increasing order of solubility in water:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}  \)
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\)
ii) In increasing order of basic strength:
a. Aniline, p - toluidine and p - nitro aniline
p - toluidine > aniline >p - nitro aniline

 b.  \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}, \mathrm{p}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2} \)
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{p}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2}\\ 2^{o} \text { amine } \quad \quad \quad \quad   e^{\ominus} \text { with drawing (group) }\)
(iii) In decreasing order of basic strength in gas phase:
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{NH},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N} \text { and } \mathrm{NH}_{3}  \)
\(\mathrm{NH}_{3}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{~N}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NH}\\ \quad \quad \quad 1^{0} \text { amine } \quad 2^{0} \text { amine } \quad \quad \quad 3^{0} \text { amine }\)
(iv) In increasing order of boiling point:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)
\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \\ 2^{0} \text { amine } \quad \quad1^{0} \text { amine }\)
Generally amines have lower boiling point than alcohol. Due to comparable molecular mass and weaker H-bonds in Amines.
(v) In decreasing order of the pK_b values:
\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \text { and } \mathrm{CH}_{3} \mathrm{NH}_{2} \)
\(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\mathrm{CH}_{3} \mathrm{NH}_{2}<\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} \)
\(\quad \quad \quad 2^{0} \text { amine } \quad 1^{0} \text { amine } 1^{0} \text { amine } \quad 2^{0} \text { amine } \)
\(\mathrm{PK}_{b}: \quad 3.00\quad < \quad 3.29 \quad < \quad 3.38 \quad<\quad 9.30\)
pK_b ,Due to + 1 effect of C₂H₅ group. Higher the value of pK_b lower is the basicity
(vi) Increasing Order of basic strength:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \text { and } \mathrm{CH}_{3} \mathrm{NH}_{2} \)
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2}>\mathrm{CH}_{3} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \\ \left(\mathrm{pK}_{6}: 9.38 \quad > \quad \quad 8.92 \quad \quad > \quad 3.38 \quad > \quad \quad 3.00\right)\)
Due to + 1 effect of C₂H₅ group.
In decreasing order of basic strength:

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.

Schottky defect:
(i) Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice.
(ii) This effect does not change the stoichiometry of the crystal.
(iii) Ionic solids in which the cation and anion are of almost of similar size show schottky defect. Ex: NaCl.
(iv) Presence of large number of schottky defects in a crystal, lowers its density.

Frenkel defect:
(i) Frenkel defect arises due to the dislocation of ions from its crystal lattice.
(ii) The ion which is missing from the lattice point occupies an interstitial position.
(iii) This defect is shown by ionic solids in which cation and anion differ in size.
(iv) Unlike Schottky defect, this defect does not affect the density of the crystal.
(v) For example AgBr, in this case, small Ag⁺ ion leaves its normal site and occupies an interstitial position.

(i) [Co(NH₃)₅ CI]²⁺
(ii) [CO(NO₂)₃(NH₃)₃]
(iii) [Co(NH₃)₄H₂OBr](NO₃)₂
(iv) [Co(en)₂CI₂]CI
(v) [Cu(NH₃)₄]SO₄

+ 2H₂O

1. \(\kappa=\mathrm{C}\left(\frac{1}{\mathrm{~A}}\right)\)
2. \(\operatorname{let} \frac{l}{A}=x\)
i) \(0.01 \mathrm{M} \mathrm{KCl}: \kappa=0.01 x=10^{-2} x\)
ii) \(0.005 \mathrm{M} \mathrm{KCl}: \kappa=0.005 x=5 \times 10^{-3} x\)
iii) \(0.1 \mathrm{M} \mathrm{KCl} \quad: \kappa=0.1 x=10^{-1} x\)
iv) \(0.25 \mathrm{M} \mathrm{KCl}: \kappa=0.25 x=2.5 \times 10^{-1} x\)
v) \(0.5 \mathrm{M} \mathrm{KCl}: \mathrm{K}=0.5 x=5 \times 10^{-1} x\)
\(\therefore 5 \times 10^{-1} x>2.5 \times 10^{-1} x>10^{-1} x>10^{-2} x>5 \times 10^{-3} x\)
\((ie) 0.5 \mathrm{M} \mathrm{KCl}>0.25 \mathrm{M} \mathrm{KCl}>0.1 \mathrm{M} \mathrm{KCl}>0.01 \mathrm{M} \mathrm{KCl}>0.005 \mathrm{M} \mathrm{KCl}\)

(i) Let us consider the reactions between a strong acid, HCI, and a weak base, NH₄OH, to produce a salt, NH₄CI, and water.
HCI_(aq) + NH₄OH_(aq) ⇌ NH₄CI_(aq) + H₂O _(I)
NH₄CI_(aq) ⟶ NH₄⁺ +CI^-_(aq)
(ii) NH₄⁺ is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with OH^- from water to produce unionised NH₄OH
NH₄⁺ _(aq) + H₂O_(I) ⇌ NH₄OH_(aq) + H⁺_(aq)
(iii) There is no such tendency shown by Cl^- and therefore [H⁺] > [OH^-]; the solution is acidic and the pH is less than 7.
(iv) The K_h and K_b are related by
\(\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}}\)
(Or)
\(\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\)
Degree of hydrolysis (h)
\(\underset{(1-h)}{\mathrm{NH}_{4}^{+}(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(j)} \rightleftharpoons \underset{h} {\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}} +\mathrm{H}^{+} \underset{h}{(\mathrm{aq})}\\ \)
\(\mathrm{K}_{\mathrm{h}} =\frac{\left[\mathrm{NH}_{4} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{NH}_{4}^{+}\right]} \\ \)
\(=\frac{\mathrm{hc} \times \mathrm{h}}{(1-\mathrm{h}) \mathrm{c}} \)
\(K_{h}=\frac{h^{2} c}{(1-h)} \)
\(\text{If } \mathrm{h}<<1 ; \mathrm{K}_{\mathrm{h}} \simeq \mathrm{h}^{2} \mathrm{c} \)
\(h^{2}=\frac{K_{h}}{c}\)
\(h=\sqrt{\frac{K_{h}}{c}} \ (or) \ h=\sqrt{\frac{K_{w}}{K_{b} \cdot C}} \quad\left(\because K_{h}=\frac{K_{w}}{K_{b}}\right)\)
Also; \(\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}} \) (or) \(\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}}\)
pH = -log [H⁺]

Considering Ba(OH)₂ to be a strong base:
\(\text { Normality } =\text { Molarity } \times \text { Acidity }  \)
\(=1.5 \times 10^{-3} \times 2\)
\({\left[\mathrm{OH}^{-}\right] } =3 \times 10^{-3}  \)
\(\mathrm{pOH} =-\log _{10}[\mathrm{OH}^-]  \)
\(=-\log _{10}\left(3 \times 10^{-3}\right)  \)
\(=-\left[\log _{10} 3+3 \log 10\right]  \)
= 3-log 3 
= 3-0.4771 
= 2.5229 
\(\mathrm{pH} =14-\mathrm{pOH} \)
= 14-2.5229 
pH = 11.4771 = 11.48

For a first order reaction
\(\\ \\ k=\frac { 2.303 }{ t } log\left( \frac { [{ A }_{ 0 }] }{ [A] } \right) \\ \)   ..(1)
Let[A₀] =100M
When 
t = t_90%; [A] = 10M (given that t_90% = 8hours)
t = t_80%; [A ] = 20M
\(k=\frac { 2.303 }{ { t }_{ 80\% } } \log\left( \frac { 100 }{ 20 } \right) \)
\({ t }_{ 80\% }=\frac { 2.303 }{ K } \log(5)\) ....(2)
Find the value of k using the given data
\(k=\frac { 2.303 }{ { t }_{ 90\% } } \log\left( \frac { 100 }{ 10 } \right) \)
\(k=\frac { 2.303 }{ 8 } \log10\)
\(k=\frac { 2.303 }{ 8 } ...(3)\)
Substitute the value of k in equation (2)
\({ t }_{ 80\% }\frac { 2.303 }{ 2.303/8hours } \log(5)\)
t_80%= 8 hours x 0.6989
t_80%= 5.59 hours

Given data: Rate constant of a first order reaction (k) = 0.45 sec^-1
Formula: \({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
Solution: \({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } =\frac { 0.693 }{ 0.45 } \)
Half-life period = 1.54 sec.

If the monomer is represented by X. Then
\(2 \mathrm{M} \rightarrow(\mathrm{M})_{2}\)
Since the reaction is of second order, the rate of reaction will be given by,
\(\text { Rate }=\mathrm{k}[\mathrm{M}]^{n}  \)
\(7.5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=\mathrm{k}\left(0.05 \mathrm{~mol} \mathrm{} \mathrm{~L}^{-1}\right)^{2} \)
\(\mathrm{k} =\frac{Rate}{[M]^{n}}  \)
\(\mathrm{k} =\frac{7.5 \times 10^{-3}}{(0.05)^{2}}  \)
\(=3 \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}\)