Solution

(a) K₃[Al(C₂O₄)₃]
(b) [Co(NH₃)₄(H₂O)Cl]Cl

Solution

Osmotic presuure=iCRT
where, C=concentration of solute in terms of Molarity)
R= Gas constant = 0.082 L(atm)(mol)⁻¹K⁻¹
T=temperature in Kelvin
i=Van't-Hoff factors=1 for non-electrolyte
5% urea solution means 5 g urea is present in 100ml of solution.
mole of urea=weight
given/Molecular weight of urea
5 g/60 g mol⁻¹= 1/12
Hence, Concentration of urea (in terms of Molarity) [C]
= (moles of urea(n) /volume of solution)×1000 ={(1/12)/100} × 1000 = 10/12
Hence Osmotic pressure=iCRT=1 × (10/12) × 0.082 × 300 atm
= 20.52 atm
or
Given,
The mass of water = 100 g
The mass of urea = 7.5 g
The mass of Z = 42.75 g
The molar mass of urea = 60 g/mol
Both the solutions have same freezing point
Since both the solutions have the same freezing point the change in freezing points (ΔT𝔣) for both the solutions is same.

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Solution

a. NH₂(CH₂)₆NH₂ – Hexamethylenediamine,
HOOC(CH₂)₄COOH – Adipic acid
b. CH₂=CH-CH=CH₂ – Butadiene,
C₆H₅CH=CH₂ – Styrene

Solution

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Solution

Solution

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Solution

(a) Because sulphur readily gets oxidized itself to more stable +6 state.

(b) Because of absence of d-orbital in Fluorine.
(c) Because size increases from Helium to Radon. Also, dispersion or van der Waal forces increase from Helium to Radon.
or
(a) MnO₂+ 4HCl —>MnCl₂ + Cl₂ + 2H₂O

(b) XeF₆+ KF —>K⁺[XeF₇]⁻

(c) I⁻ (aq) + H⁺ (aq) + O₂(g) —>2I₂(s) + 2H₂O(l)

Solution

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or

(a) (i) Add Ice cold (NaNO₂ + HCl) followed by phenol or β-Naphthol to both the compounds.
Aniline forms orange red dye while ethylamine doesn’t.
ii) Add CHCl₃ and KOH (alc.) to both the compounds.
Aniline gives foul smelling isocyanide while N-Methylaniline doesn’t.
(b) CH₃- CH₂- CH₂ - CH₂– OH (butanol)
CH₃-CH₂-CH₂- CH₂- NH₂ (butanamine)
CH₃- CH₂- CH₂- CH₃ (butane)
The hydrogen bonding is strongest in butanol thus the decreasing order of boiling point is:
Butanol > Butanamine > Butane

Solution

(a) Because the – CHO group in glucose is involved in hemiacetal formation and thus is not free i.e., due to cyclic structure of glucose -CHO group is not free.

(b) The two strands of DNA are bound together by hydrogen bonds between the purine base on one strand and the pyrimidine base on the other strand, and vice versa. The only possible pairings in DNA are G (guanine) and C (cytosine) through three H-bonds (i.e., C≡≡G)) and A (adenine) and T (thymine) through two H-bonds (i.e., A = T). This is due to the different sizes and geometries of the bases. The sequence of bases in one strand fixes the sequence of bases in the other strand due to this base-pairing concept. The two strands are thus complimentary rather than identical.

(c) Starch consists of amylose and amylopectin.
● Amylose is a long unbranched chain with a-D(+)-glucose units that are held together with C1–C4 glycosidic linkage.
●Amylopectin is a highly branched chain polymer of a-D-glucose units in which glycosidic linkage is present between C1–C4 atoms and branching occurs through C1–C6 glycosidic linkage.
Cellulose is a straight chain polysaccharide composed only of b-D-glucose units which are joined by glycosidic linkage between C1 of one glucose units and C4 of next glucose units.

Solution

(a) If the surface area of a reactant is increased: more particles are exposed to the other reactant. there is a greater chance of particles colliding, which leads to more successful collisions per second, hence the rate of reaction increases.

(b) On decreasing the temperature, the kinetic energy of the molecules becomes smaller. There are smaller and less effective collisions between the molecules. This results in decrease in the rate of reaction.

(b) Catalysts typically speed up a reaction by reducing the activation energy. The presence of a catalyst increases the reaction rate (in both forward and reverse reactions) by providing an alternative pathway with lower activation energy.