(i) The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature and the molarity of the solution is used instead of molality.
(ii) Oxygen is present in dissolved state in water. As per Henry's law when temperature rises, solubility of a gas decreases in solvent, it means solubility of oxygen in warm water is less than cold water. This makes aquatic species respirate comfortably in cold water.
(iii) Elevation in boiling point is directly proportional to 'i'. ΔTb ∝ i. Now as given in the question, elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution. It is because KCl being ionic, dissociates into K⁺ and CI^- and therefore it's van't Hoff factor, i is 2 whereas for sugar van't Hoff factor is 1 as it does not undergoes such a dissociation

symmetry of the molecule is related to its crystallattice structure and hence, to its melting point and solubility. A highly symmetrical structure has a higher melting point
The melting point of p-isomer of dichlorobenzene is higher than that of o- and m-isomers. This is because p-isomer has symmetrical structure due to which its molecules can easily pack closely in crystal lattice. Thus, it has stronger intermolecular forces of attraction than o- and m-isomers.
Difference lies in the mode and extent of dissociation of KOH in the presence of different solvents OH^- is a good nucleophile while C₂H₃O^- is a strong base. In aqueous solution, KOH is almost completely ionised to give OH- ions which being a strong nucleophile gives a substitution reaction on alkyl halide to form alcohol.
But an alcoholic solution of KOH contains alkoxide ions RO^- which being much stronger base than OHions preferentially eliminates a molecule of HCI from an alkyl chloride to form alkenes.

(i) Sum of enthalphy of atomisation and ionisation is greater than enthalphy of hydration. Thus, \(E_{(M^{2+}/Mn)}^0\) is positive for copper.
(ii) Ionisation enthalpy is lower than hydration enthalphy due to stable 3d⁵ configuration. Thus, \(E_{(M^{2+}/Mn)}^0\) is more negative.
(iii) Cr²⁺ is a stronger reducing agent because it can lose electron to form Cr³⁺ which has stable 3d³ configuration (half-filled t_2g³). Further is more negative than Fe²⁺

Given, W(cane sugar) = 5 g
W(X) = 0.877 g
M(cane sugar) = 342 g mol^-1
\(\pi (cane \ sugar)= \pi(X)\)
\(\frac{W(cane\ sugar)\times 1000}{M(cane \ sugar) \times V}= \frac{W(X)\times 1000}{M(X)\times V}\)
\(\frac{5 g}{342 g}=\frac{0.877 g }{M(X)}\)
M(X) = 59.9 g mol^-1 or 60 g mo^-1.

(i)

(ii) CH₃CH₂CI + AgNO₂\(\rightarrow\)CH₃CH₂NO₂+AgCI
(iii) CH₃CH₂CH₂CH(Br)CH₃ + KOH(alc.)\(\rightarrow\)CH₃CH₂CH=CH CH₃

(i) Chlorobenzene to biphenyl: When two chlorobenzene combine with sodium metal in the presence of dry ether it forms biphenyl.
(ii) Propene to 1-iodopropane :
CH₃CH = CH₂ + HI \(\overset { Peroxide }{ \underset { Anti-Markovnokov's\ rule }{ \longrightarrow } } \)CH₃CH₂I + CH₃CHICH₃
                                                                    (Major)        (Minor)
CH₃CH = CH₂\(\overset { HBr/Peroxide }{ \longrightarrow } \)CH₃CH₂CH₂Br\(\overset { NaI/acetone }{ \longrightarrow } \)CH₃CH₂CH₂I
(iii) 2 - bromobutance to but-2-ene
\({ H }_{ 3 }C-{ CH }_{ 2 }-\underset { \overset { | }{ Br } }{ CH } -{ CH }_{ 3 }\overset { Alc.KO }{ \longrightarrow } \quad { H }_{ 3 }C-\underset { But-2-ene\\ \quad \quad (80%) }{ CH } =CH-{ CH }_{ 3 }+{ CH }_{ 3 }-\underset { But-1-ene\\ \quad \quad (20%) }{ { CH }_{ 2 } } -CH={ CH }_{ 2 }\)

(i) Dispersion forces
(ii) Dispersion forces
(iii) Ion-dipole attraction
(iv) H-bound and dipole-dipole attraction
(v) Dipole -dipole attraction.

(i) Isotonic solutions : An isotonic solution refers to two solutions having the same osmotic pressure across a semipermeable membrane. This state allows for the free movement of water across the membrane without changing the concentration of solutes on either side.
(ii)Hypertonic solution: A hypertonic solution is a particular type of solution that has a greater concentration of solutes on the outside of a cell when compared with the inside of a cell.
(iii) Hypotonic solution: The solution having lower osmotic pressure than another solution.

With rise in temperature, it loses water of crystallization.

\(\underset { Methyl\\ Chloride }{ CH_{ 3 }CI } +AgNO_{ 2 }+AgCI\)

(i) It is due to presence of unpaired electrons in d-orbitals therefore, they undergo d-d transitions by absorbing light from visible region and rediate complementary colour.
(ii) It is because neither Zn nor Zn²⁺ ion has incompletely filled d-orbital.

(i) It is because these metals show variable oxidation states and form unstable intermediate which readily change into product.
(ii) It is due to less energy required for more number of electrons to take part in metal-metal bonding in 4d and 5d series.

\(\pi =CRT,i.e.,\pi \alpha C\)

Urea and glucose do not dissociate. NaCI dissociates to give 2 particles.

In a ideal solution, solute does not undergo dissociation or association. Hence, van't Hoff factor = 1.

As solute is solid, the solvent, i.e., water will freeze out.

(a) (i) Ideal solution: Those solutions which follow Raoult's law at specific temperature over the entire range of concentration are called Ideal solution.
(ii) Azeotrope: A liquid mixture which distills at constant temperature without under going any changes in composition is called Azeotrope.
(iii)Osmotic pressure: The minimum excess pressure that has to be applied on the solution side to prevent the entry of the solve t into the solution through the semi-perable membrane is called osmotic pressure.
(b) Given : Mass of solute, \(\omega \) = 10 g
Mass of solvent, W = 90 g
Molar mass of solute, M = 180
\(Molality=\frac { \omega \times 1000 }{ M\times W } \)
\(=\frac { 10\times 1000 }{ 90\times 180 } \)
= 0.6172 \(\approx \) 0.62 mol kg^-1

(i) Mn³⁺ (3d4) on changing to Mn²⁺ (3d5) becomes stable, half filled configuration has extra stability. Therefore, Mn³⁺ can be easily reduced and acts as a good oxidizing agent.
(ii) EO(M₂+/M) values are not regular in the first transition series metals because of irregular variation of ionization enthalpies (IE₁ + IE₂) and the sublimation energies.
(iii) Among transition elements, the bonds formed in +2 and +3 oxidation states are mostly ionic. The compounds formed in higher oxidation states are generally formed by sharing of d-electrons. 
(a) The transition elements show variable oxidation states because their atoms can lose a different number of electrons. This is due to the participation of inner (n - 1) d-electrons in addition to outer electrons because the energies of the ns and (n - 1) d-subshells are almost equal.
(i) Manganese
(ii) Scandium
(b) The steady decrease in atomic and ionic sizes of lanthanide elements with increasing atomic number is called lanthanide contraction. In the lanthanoids, there is a regular decrease in the size of atoms and ions with an increase in atomic number. For example, the ionic radii decrease from Ce³⁺(111 pm) to Lu³⁺ (93 pm). Cause of lanthanoid contraction. As we move through the lanthanoid series, 4f-electrons are being added, one at each step. The mutual shielding effect of electrons is very little, even smaller than that of d-electrons. This is due to the shape of f-orbitals.
The nuclear charge, however, increases by one at each step. Hence, the inward pull experienced by the 4f-electrons increases. This causes a reduction in the size of the entire 4f' shell. The sum of the successive reductions gives the total lanthanoid contraction.
The important alloy is misch metal.

\(\underset { n-Butylchloride }{ { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 2 }CI } +KOH(alc.)\overset { \triangle }{ \longrightarrow } \underset { But-l-ene }{ { CH }_{ 3 }{ CH }_{ 2 }CH={ CH }_{ 2 }+KCI+{ H }_{ 2 }O } \)

\((iv)\quad { 2CH }_{ 3 }Br+2Na\overset { Dry\quad ether }{ \underset { \quad Wurt\\ reaction }{ \longrightarrow } } \underset { Ethane }{ { CH }_{ 3 }{ CH }_{ 3 } } +2NaBr\)
\(\\ (vi)\quad \underset { Methyl\\ chloride }{ { CH }_{ 3 }CI } +KCN\longrightarrow \underset { Methyl\\ chloride }{ { CH }_{ 3 }CN } +KCI\)

C) Assertion is correct statement but reason is wrong statement.
Explanation:
Benzene and toluene do not form H-bonding with each other.

(a): A half-filled or fully-filled orbital is more stable than incompletely filled orbital.

(a) It is because neither they nor their ions have incompletely filled d-orbitals.
(b) Scandium (Sc) and Zinc (Zn).
(c) It is because they show variable oxidation state, can form intermediate complexes and have large surface area for adsorption of gases.
(d) It is due to strong interatomic forces of attraction due to presence of unpaired electrons.
(e) It is because Cu^{2 +} has one unpaired electron and undergoes d-d transition by absorbing light from visible region and radiate blue colour, where as Zn^{2 +} is colourless due to absence of unpaired electron.

(i) (b) : Moles of C₆H₆ = \(\frac{7.8}{78}=0.1\)
Mole C₆H₅CH₃ = \(\frac{9.2}{92}=0.1\)
Mole fraction of C₆H₆ = \(\frac{0.1}{0.1+0.1}=0.5\)
=> Mole fraction of C₆H₅CH₃ = 0.5
Vapour pressure of toluene = Vapour pressure of pure toluene x mole fraction of toluene
= 0.0925 x 0.5 = 0.04625
Vapour pressure of benzene = 0.256 x 0.5 = 0.128
Total vapour pressure of solution = 0.17425
(ii) (a) : Moles of benzene in solution-II = \(\frac{3.9}{78}=0.05\)
Moles of toluene in solution-II = \(\frac{13.8}{92}=0.15\)
Vapour pressure of solution
= 0.256 x 0.05 + 0.0925 x 0.15
= 0.0128 + 0.013875 = 0.026675
(iii) (c) : Mole fraction of benzene in vapour phase
\(y_{\text {benzene }}=\frac{p_{\text {benzene }}}{P_{\text {total }}}=\frac{0.128}{0.17425}=0.734\)
(iv) (a) : Benzene and toluene form an ideal solution.