Answer:
coagulation     (1)

Answer:
[Co(NH₃) ₄(H₂O) ₂]Cl₃     (1)

Answer:
Due to intermolecular H bonding in alcohols, which is absent in ethers, alcohols have a higher boiling point than ethers.     (1)

Answer:
5-Formyl-3-methylpentanoic acid     (1)

Answer:
Due to electron releasing inductive effect of alkyl group in aliphatic amines, electron density on N atom increases, making them stronger bases than ammonia.     (1)

Answer:
The isomers which differ in configuration at the hemiacetal carbon i.e. anomeric carbon are called anomers. In case of glucose anomers differ in orientation of hydroxyl group at C-1. Eg. α-D-Glucopyranose and β-D- Glucopyranose.     (1)

Answer:
COOH(CH₂) ₈COOH and NH₂ (CH₂) ₄NH₂.     (1)

Antiseptics	Disinfectants
Chemical substances which either kill microorganisms or prevent their growth and are not harmful to the living tissues are called antiseptics. Being unharmful they can be applied to living tissues.	Chemical substances which either kill microorganisms or prevent their growth and are harmful to the living tissues are called disinfectants. Being harmful they cannot be applied to living tissues.
Eg. 0.2% phenol solution	Eg. 1% phenol solution
0.5	0.5

Answer:
ΔT_f =i K_fm     (.5)

For KCl, i = 2
Molar mass=74.5gmol^-1
m = ΔT_f /iK_f = 3/2X1.86 = 0.81 mol     (0.5)

Mass of KCl required = 0.81X 74.5= 60.345g     (1)

Answer:
At cathode: O₂ (g) + 2 H₂O(l) +4e^- à 4OH- (aq)     (1)
At anode: 2H₂ (g) + 4OH^- (aq)à 4H₂O (l) + 4 e^-     (1)

Answer:

2K2CrO4 + H2SO4 (conc)	à	K2Cr2O7+ K2SO4 + H2O
B (Yellow)		A (orange)

4 K2Cr2O7 A Potassium dichromate

heat à

4 K2CrO4 B Potassium Chromate

+

2 Cr2O3 C Chromium trioxide

+

3O2

K2Cr2O7 A

+ H2SO4 + SO2 --------à K2SO4 +

Cr2 (SO4) 3 D(Green)

+ H2O

(0.5 mark for correctly identifying each A, B,C,D)

Answer:
When ferrous sulphate solution is mixed with ammonium sulphate in 1:1 ratio, a double salt called Mohr’s salt is formed FeSO₄ (NH₄) ₂SO₄.6H₂O. Double salt ionizes in water giving Fe²⁺ ions. But when CuSO₄ is mixed with aqueous ammonia, there is formation of a coordination complex [Cu(NH₃) ₄]SO₄, which does not give Cu²⁺ ions in solution.     (2)

Answer:
ccp is formed by B, thus number of B = 8X1/8=1     (0.5)
Number of octahedral voids= No. of B = 1

Number of tetrahedral voids= 2xNo. of B = 2

No. of A = 1+ 2x1/2= 2 (0.5)
Formula of the compound is BA₂     (1)

OR

For bcc, number of atoms per unit cell = 2
No. of atoms in 3.9g of K = 3.9x6.022x10²³/39= 6.022x10²²     (1)
No. of unit cells in 3.9 g of K = 6.022x10²²/2 = 3.011x10²²     (1)

Answer:
Let % of Fe²⁺ be x
            % of Fe³⁺ be (0.93- x )
Total charge on cations is equal to total charge on anions.

2x + 3(0.93-x) = 2     (0.5) 2x + 3(0.93-x) - 2 =0

x= 0.79 (0.5)

% of Fe²⁺ = = 84.9%     (0.5)
% of Fe³⁺ = 15.1%     (0.5)

Answer:
Polyacrylonitrile is used as a substitute for wool.     (1)
Its monomer is acrylonitrile, CH2=CHCN     (1)

Answer:
The drugs which counteract the effect of histamines released due to allergic reactions are antihistamines. Eg. Chlorpheniramine (1)
They compete with histamines for binding to the receptor at the site where histamine exerts its effect, hence they interfere with normal action of histamines.     (1)

Answer:
C₆H₅CH₂Br<C₆H₅CH(CH₃)Br< C₆H₅CH(C₆H₅)Br< C₆H₅C(CH₃)( C₆H₅)Br.(SN1)     (1)
C₆H₅C(CH₃)( C₆H₅)Br< C₆H₅CH(C₆H₅)Br < C₆H₅CH(CH₃)Br <C₆H₅CH₂Br .(SN2)     (1)

Answer:
When chloroethane reacts with KCN, propanenitrile is formed with liberation of potassium chloride
CH³CH₂Cl + KCN à CH₃CH₂CN + KCl     (1)
When 1-bromopropane reacts with silver acetate, propyl ethanoate and silver bromide are formed
CH₃CH₂ CH₂Br + CH₃COOAg --à CH₃COOCH₂CH₂CH₃ + AgBr     (1)

Answer:
Boiling point is the temperature at which vapour pressure is equal to atmospheric pressure. In a solution containing non volatile solute, vapour pressure is lowered as surface of liquid is occupied by some molecules of non volatile solute also. Hence high temperature is required to boil it.     (1)

    (1)

Answer:

Answer:
(a) The process of converting a freshly prepared precipitate into a colloid in the presence of a peptizing agent is called peptusation. During this process, the precipitate adsorbs one of the ions and gets converted to a colloid.     (1)

(b) Activated charcoal adsorbs coloured impurities present in solution thus the remaining solution is colourless.     (1)

(c) The substances which behave as strong electrolyte at low concentration. And at higher concentration behave as colloid due to formation of aggregates is called associated colloids.
Eg. Soap solution.     (1)

Answer:
(a) 3CaO + 2 Al ----à Al₂O₃ + 3Ca
Δ_rG⁰= Δ_rG⁰Al₂O₃ -3ΔfG0CaO     (1)
= -1582.4 -3x(-604.2) = 230.2 KJmol-1

since ΔrG0 is positive, reduction reaction is not feasible. (1)

(b) The principle behind zone refining is that impurities are more soluble in molten metal than in solid metal.     (1)

Answer:

1.  
   1. Ammonia has lone pair on N atom which is easily available for donation due to small size and high electronegativity of N.     (1)
   2. N exibihits oxidation states like +2,+3,0,-3 etc. in HNO₂, N has an oxidation state of +3. It can change to a lower state, thereby acting as an oxidizing agent and to a higher state therby acting as a reducing agent also.     (1)
2. XeF₆ on complete hydrolysis yields xenon trioxide and hydrogen fluoride. XeF₆ + 3 H₂O -à XeO₃+ 6HF     (1)

OR

Due to small size of fluorine, bond length of F-F bond is less than Cl-Cl bond which leads to high interelectronic repulsions between the lone pairs of electrons. Thus bond dissociation energy is less.     (1)

O because of its small size is capable of forming Pπ - Pπ bond and forms O₂ .Since intermolecular forces involved are weak van Der Waals forces, it is a gas.S being large in size is not able to form bond and thus does not form S₂ molecule. It rather exist as rings or chain in which atoms are linked by comparatively stronger intermolecular forces as compared to that present in O₂ .Hence oxygen is a gas while sulphur is a solid     (1)

XeOF4 is isostructural with IF₅. The molecules are square pyramidal.     (1)

Answer:
(i) Transition elements form small cations with high charge density. They have vacant d orbitals available for coordination, thus they form complexes.     (1)
(ii) A steady decrease in size of elements of lanthanoid series with increasing atomic number is called lanthanoid contraction. Lanthanoid contraction is due to intervention of 4f orbital which must be filled before 5d series of elements begins.     (1)
Cr contains 6 unpaired electrons thus resulting in strong Cr-Cr bond. Hence high melting point.     (1)

Answer:
(i)
(a)

Experiment	Observation	Sample
Add NaNO2 and conc. HCl . cool to 0°C. Add alkaline β-naphthol solution	Orange dye is formed	. Aniline (0.5 )
Add NaNO2 and conc. HCl . cool to 0°C. Add alkaline β-naphthol solution	No dye formed	Benzyl amine (0.5)

(ii)

Experiment	Observation	Sample inference
Add nitrous acid(NaNO2 + HCl)	N2 gas evolved,effervescence observed	Ethyl amine (0.5 )
Add nitrous acid(NaNO2 + HCl)	Yellow oily product formed	Dimethyl amine (0.5)

(b) A reaction where acid amides are made to react with bromine and aqueous, or ehthanolic KOH to yield primary amines with one carbon less than the corresponding amide is Hoffman Bromamide reaction. It is used in stepping down the series.
    (0.5)

Answer:
(a) Glucose does not restore the colour of Schiff’s reagent and also does not react with NaHSO3.     (1)
(b) Due to presence of acidic -COOH group and basic -NH2 group, amino acids occur in zwitter ion form, thus are water soluble.     (1)
(c) DNA contains cytosine and thymine     (0.5)
RNA contains cytosine and uracil     (0.5)

Answer:
(a) Mechanism need to be added

Step 1:

Step 2:

Step 3:
    (2)

(b) CH3CH2OCH2CH3     (1)

Answer:

Answer:

Answer: