Allylic.

68% HNO₃ means that 68 g HNO₃ is present in 100 g of solution.
Moles of ​​​​​​ ​\(HNO_{ 3 }=\frac { 68 }{ 63 } =1.08\)
Density of solution \(=1.504g \ mL^{ -1 }\)
Volume of solution \(=\frac { 100 }{ 1.504 } =66.49\)
Molarity\(=\frac { 1.08 }{ 66.49 } \times 1000=16.24 M\)

(i)

(ii)
 
(iii)

(i)

(ii)
 
(iii)
 
(iv)

34.2 g/litre

Here,  \({ K }_{ H }=4.27\times { 10 }^{ 5 }mm,\ p=760 \ mm\)
Applying Henry's law, \(p={ K }_{ H }x, we \ have \ x=\frac { p }{ { K }_{ H } } =\frac { 760 \ mm }{ 4.27 \times{ 10 }^{ 5 }mm } =1.78 \times{ 10 }^{ -3 }\)
i.e., mole fraction of methane in benzene = 1.78 x 10^-3

Dichloromethane has highest dipole moment among CH₂​Cl₂​, CHCl₃​ and CCl₄​. The decreasing order of dipole moments is CH₂​Cl₂​>CHCl₃​>CCl₄​. These molecules have tetrahedral geometry due to sp3 hybridization of carbon atom.In CCl4​, the individual C−Cl bond dipoles cancel each other which results in zero dipole moment. 
Hence, CCl₄​ is non polar.

(i) As cyclohexane and n-octane both are non-polar. Hence, they will mix completely in all proportions. 
(ii) KCl is an ionic compound while n-octane is non-polar. Hence, KCl will not dissolve in n-octane. 
(iii) CH₃OH and CH₃CN both are polar but CH₃CN is less polar than CH₃OH. As the solvent n-octane is non-polar, CH₃CN will dissolve more than CH₃OH  in n-octane. Therefore, the order of solubility will be KCl 3OH < CH₃CN < cyclohexane.

A hydrocarbon with the molecular formula, C₅H₁₀ belongs to the group with a general molecular formula CnH2n. Therefore, it may either be an alkene or a cycloalkane. Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should be a cycloalkane. Further, the hydrocarbon gives a single monochloro compound, C₅H₉Cl by reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the hydrocarbon must contain H−atoms that are all equivalent. Also, as all H−atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. 

15 ppm corresponds to 15g chloroform in 1000,000g  of solution.
(i) Percent by mass =\(=\frac{\text { Mass of chloroform }}{\text { Mass of solution }} \times 100\)
Percent by mass =\(\frac{15}{1000,000} \times 100=1.5 \times 10^{-3} \)
(ii) Molality =\({\text { Mass of chloroform }}{\text { Molar mass of chloroform } \times(\text { Mass of solution - mass of chloroform) }} \times\)1000g (all masses in g)
Molality=\(\frac{15}{119.5 \times(1000,000-15)}\)\(\times 1000=1.255 \times 10^{-4} \mathrm{~m}\)
Molality = \(\frac { 15/119.5 }{ { 10 }^{ 6 } } \times 1000=1.25\times { 10 }^{ -4 }m\)

(a) (CH₃ ) C-I will be more reactive because C-I has lower bond dissociation enthalpy than C-Br bond, due to longer bond length.
(b) It is because they do not differ appreciably in their boiling points (physical properties), differ in optical rotation and biological properties.
(c)  because it is primary halide and has less stearic hinderance.
(d)  is more reactive because -NO₂ being electron withdrawing stabilizes intermediate negatively charged ions.
(e)  is optically active because it has chiral 'C' atom (asymmetric carbon).

Given that, total pressure of air in equilibrium with water = 10 atm
As air contains 20% oxygen and 79% nitrogen by volume.
ஃ Partial pressure of oxygen
(\(\rho\)\(o_{2}\)) = \(\frac{20}{100}\) x 10 atm = 2 atm = 2 x 760 mm = 1520mm
Partial pressure of nitrogen, (\(\rho\)_{\(N_{2}\)} ) = \(\frac{79}{100}\)x 10 atm
 = 7.9 atm = 7.9 x 760 mm = 6004 mm
Given that, K_H(O₂) = 3.30 x 10⁷ mm,
K_H(N₂) = 6.51 x 10⁷ mm
According to Henry's law,
\(\rho\)\(o_{2}\)= K_H \(\times\) \(\chi\)\(o_{2}\)
or \(\chi\)\(o_{2}\) =\(\rho\)\(o_{2}\) /K_H  =\(\frac{1520 \ mm}{3.30\times 10^7\ mm}\) = 4.61 x 10^-5
\(\rho\)_{\(N_{2}\)} = K_H x \(\chi\)_{\(N_{2}\)}
or  \(\chi\)_{\(N_{2}\) = \(\rho\)\(N_{2}\)}/K_{H =\(\frac{6004 mm}{6.51* 10^7mm}\)}= 9.22 x 10^-5

C) Assertion is correct statement but reason is wrong statement.
Explanation:
Elevation in boiling point and depression in freezing point are colligative properties because they depend only on the number of solute particles in a solution irrespective of their nature. But all colligative properties are not used for mass determination for every substance, as some give very low precision, or require excessive heating or cooling.

(b) (A) is not correct but (R) is correct. In fact, the use of thionylchloride is preferred over phosphorus chlorides for preparing haloalkanes from alcohols because the other products of the reaction (i.e.) HCI(g) and SO₂(g) being gases escape out leaving behind pure haloalkane.