Molality does not change with temperature while molarity changes with temperature.

\({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ CH } -{ CH }_{ 2 }Br+KOH(a/c)\longrightarrow { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ CH } ={ CH }_{ 2 }\overset { HBr }{ \longrightarrow } { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ Br } }{ C } } -{ CH }_{ 3 }\)
             'A'                                                                            'B'                                          'C'
'C' is the isomer of 'A'.
\(2{ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ CH } -{ CH }_{ 2 }Br+2Na\overset { Dry }{ \underset { ether }{ \longrightarrow } } { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ CH } -{ CH }_{ 2 }-{ CH }_{ 2 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ CH } -{ CH }_{ 3 }+2NaBr\)
\(\\ 2{ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 2 }Br+2Na\overset { Dry }{ \underset { ether }{ \longrightarrow } } { CH }_{ 3 }-{ CH }_{ 2 }-\overset { 'D'(C_{ 8 }{ H }_{ 18 }) }{ \underset { n-Octane(C_{ 8 }{ H }_{ 18 }) }{ { CH }_{ 2 }-{ CH }_{ 2 }-{ CH }_{ 2 } } } -{ CH }_{ 2 }-{ CH }_{ 2 }-{ CH }_{ 3 }\)
'D' and n-octane are different.

It is because alcohol can form H-bonds with H₂O, whereas hydrocarbons cannot, that is why alcohols are soluble in water.

We know that, m = K_H \(\times\) P
\(\therefore\) 6.56 \(\times\) 10^-2g = K_H \(\times\) 1 bar ...(i)
\(\therefore\) 0 5.00 \(\times\) 10^-2 g = K_H \(\times\) P.........(ii)
K_H = 6.56 \(\times\) 10^-2/1 bar (from i)
K_H = 5.00 \(\times\) 10^-2//p bar (from ii)
\(\therefore \quad \frac{6.56 \times 10^{-2}}{1}=\frac{5.00 \times 10^{-2}}{p}\)
\(\therefore \quad P=\frac{5 \cdot 00}{6 \cdot 56}=0.762 \text { bar. }\)

The egg placed in pure water will swell, whereas the egg placed in saturated solution of NaCl will shrink.

\(\pi=C R T=\frac{n}{V} R T=\frac{W_{2} \times R T}{M_{2} \times V}\)
For both solutions, R, T and V are constant.
ஃ In the first case, (M₂ = 180 for glucose)
\(98=\frac{36}{180} \times \frac{R \times 300}{V}\)              ----(i)
In the second case,
\( 1.52=\frac{W_{2}}{M_{2} V} \times R \times 300\\ \)  ------(ii)
On dividing Eq. (ii) by Eq. (i), we get
\(\frac{W_{2}}{M_{2} V} \text { or } C=0.061 \mathrm{~mol} \mathrm{~L}^{-1}\)

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. 

\(\pi =CRT\\ C=0.002\quad mol\quad { dm }^{ -3 }=2\quad mol\quad { m }^{ -3 }\\ R={ 8.314JK }^{ -1 }{ mol }^{ -1 },T=293K\\ \pi =2\times 80314\times 293=4872\quad Pa\)

\(NaCl\longrightarrow { Na }^{ + }+{ Cl }^{ - },i=2\\ \Delta { T }_{ f }={ iK }_{ f }m=2\times 1.86\times 0.1={ 0.372 }^{ o }C,\\ { T }_{ f }={ 0.372 }^{ o }C.\)

(i) Since compound (A) on treatment with CHCI3 and KOH (i.e., Reimer-Tiemann reaction), gives two products Band C, therefore, A must be phenol and Band C must be o-hydroxybenzaldehyde and P: hydroxybenzaldehyde respectively or vice-versa.
(ii) Since both Band C on distillation with Zn dust give the same compound (D), therefore, D must be benzaldehyde.
(iii) Since oxidation of D gives E with M.F. C7R602, therefore, E must be benzoic acid.
(iv) Since sodium salt of E, i.e.benzoic acid upon heating with soda-lime gives compound (F), therefore.
(F) must be benzene. Zn dust distillation of A would also give benzene (F).

(b) Both (A) and (R) are true but (R) is not the correct explanation of (A). The correct explanation is that one molar aqueous solution has 1 mole in less than 1000 g of water. Hence, 1000 g of water will be associated with more than 1 mole while 1 molal has 1 mole in 1000 g of water

(d) Lowering of vapour pressure is directly proportional to the number of species present in solution. Sugar is a non-electrolyte however KCI is an electrolyte and dissociates in solution to give particles. Hence, there is greater lowering in vapour pressure in case of KCl. Hence, the vapour pressure of 0.1 M sugar solution is more than that of 0.1M potassium chloride solution. Thus, Assertion (A) is incorrect but Reason (R) is correct.

i. (c) phenol
ii. (d) 9
iii.(b) Kolbe’s reaction
iv.(d) as an intestinal antiseptic.