Some liquid molecules at the surface are replaced by the molecules of the solute which are non - volatile.

CH₃−CH(Cl)CH₂CH₃ will be more easily hydrolysed because it will form secondary carbocation which is more stable than primary carbocation.

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

It is because forces attraction between ethyl alcohol and water are less than ethanol-ethanol and water-water. It shows positive deviation.

Solubility of Solids In Liquids:
Solubility is the interaction between particles of the solute and the solvent. Hence, it depends on the nature of the solute and the solvent, as well as on the temperature and pressure. But the solubility of solids in liquids is independent of pressure.
Factors affecting solubility
1. Temperature. Basically, solubility increases with temperature. 
2. Polarity. In most cases solutes dissolve in solvents that have a similar polarity.
3. Pressure. Solid and liquid solutes.
4. Molecular size. 
5. Stirring increases the speed of dissolving.

(a) (i) Molarity: Molarity of a substance in a solution is equal to the number of moles of the substance present in one litre of the solution.
i.e., Molarity = \(\frac { No.\ of\ moles\ of\ substance }{ Volume\ of\ solution\ in\ L } \)
(ii) Molal elevation constant (K_b): It is also called ebullioscopic constant. It is equal to the change in boiling point of one molar solution.
\(\Delta\)T_b = K_b . m
So, when, Molarity = 1,\(\Delta\) T_b = K_b
(b)Given; Mass of urea, W_B = 15 g
Molar mass of rea, M_B = 60 g
The solution of urea in water is isotonic to that of glucose solution.
So,  \(\pi _{ urea }=\pi _{ Glucose }\)
\(C_{ urea }RT=C_{ Glucose }RT\)
\(\frac { { n }_{ urea } }{ V } RT=\frac { { n }_{ Glucose } }{ V } RT\)
\(=\frac { 15 }{ 60 } =\frac { { W }_{ Glucose }V }{ 180 } \)
\(={ W }_{ Glucose }=\frac { 18\times 18 }{ 60 } \)
= 45 g
So, 45 g of glucose is present in 1 L of solution.

tert-Alkyl halides prefer to undergo dehydrohalogenation in presence of a strong base such as Na metal instead of undergoing Wurtz reaction as explained below:
\(\underset{tert-Butyl\ bromide}{(CH_3)_3C-Br}+2Na\longrightarrow\underset{tert-Butylsodium}{(CH_3)_3Na^+}+Na^+Br^-\)
Thus, only rand 2° alkyl halides undergo Wurtz reaction while 3° alkyl halides prefer to undergo dehydrohalogenation to form alkenes.

The reactions are given above:
(i) 1-methylcyclohexene is the major product.
Only one alkene is formed at beta H atoms on either side of Br atoms are equivalent.

(ii) 2-methyl-2-butene is the major product.
Two different alkenes are formed. More substituted alkene is the major product as it is more stable. This is in accordance with Saytzeff's rule.

(iii) 3,4,4-Trimethylpent-2-ene is the major product.
Two different alkenes are formed. More substituted alkene is the major product as it is more stable. This is in accordance with Saytzeff's rule.

(a) Osmosis is the flow of solvent from solution of lower concentration to higher concentration through a semi-permeable membrane.
Osmotic pressure is the excess pressure which must be applied to a solution to prevent the passage of solvent through a semi-permeable memberane.
It has been found experimentally that for n moles of the solute dissolved in V litres of the solution, the osmotic pressure (π) at temperature T is
πV=nRT
Where R is a gas constant.
or \(\pi={n\over V}RT\)
= C RT
Where Cis the molar concentration ofthe solution.For a solution at given tempeature, both R and T are constant, so that
π ∝ C
Thus, osmotic pressure depends upon the molar concentration ofsolution and therefore, is a colligative property.
(b) \(ΔT_b={iK_b\times1000\times W_2\over W_1\times M_2}\)
NaCI dissociates as:
NaCI ⇾ Na⁺ + Cl^-
i = 2
W₂ = 1.5.0g, W₁ = 250.0 g, M₂ = 58.44 g mol^-1 K_b = 0.512 K kg mol^-1
\(\Delta T_b={2\times 0.512\times1000\times15.0\over 250.0\times58.44}\)
= 1.05 °C
Boiling point of solution = 100 + 1.05
= 101.5° C

Haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions because of the following reasons:
1. Resonance effect: In haloarenes, there is delocalisation of electrons due to resonance. For example, chlorobenzene is considered to be a resonance hybrid of the following structures:

It is evident that, the contribution of structures III, IV and V imparts a partial double bond character to the carbon-chlorine bond. This is confirmed by X-ray analysis which shows that the C - CI bond length in chlorobenzene is 1.69A while the C-CI bond length in ethyl chloride molecule is 1.82A. The shortening of bond length imparts stability to aryl halides and the bond cleavage becomes rather difficult. The aryl halides are, therefore, less reactive than alkyl halides.
2. Different hybridisation states of carbon atom: In haloalkanes the carbon atom of the C - X bond is sp³ hybridised while in haloarene halides, the carbon atom is sp² hybridised. The Sp² hybridised carbon atom with a greater s-character is more electronegative. It can hold the electron pair of the bond more tightly than the Sp³ hybridised carbon atom in alkyl halides. Therefore, it has less tendency to release electrons to the halogen. As a result, the bond cleavage in aryl halides is some what more difficult than in alkyl halides.

Thus haloarenes are less reactive towards the substitution reactions than haloalkanes.

Sodium chloride \((NaCI)\) is preferred over calcium chloride \(({ CaCI }_{ 2 })\).This is because less amount of sodium chloride is needed in order to bring about same depression in freezing point. For example, if it is desired to decrease the freezing point by 10°C,the amount of NaCI d CaCl2 needed to be dissolved in
1 kg of water can be calculated from the formula:
\(\triangle { T }_{ f }=\frac { { iK }_{ f }\times { W }_{ B }\times 1000 }{ { W }_{ A }\times M_{ B } }\)
\(ForNaCI,\quad NaCI\rightleftharpoons { Na }^{ + }{ CI }^{ - }\quad i=2\)
\(\triangle { T }_{ f }={ 10 }^{ \circ },{ W }_{ A }=1000g,{ K }_{ f }=1.86 \ K{ m }^{ -1 },{ W }_{ B }=?,\)
\( { M }_{ B }=58.5\)
\( 10=\frac { 2\times1.86\times{ W }_{ B }\times1000 }{ 1000\times58.5 } \)
\( or \ \ \ \ { W }_{ B }=\frac { 10X1000X58.5 }{ 2X1.86X1000 } =157.26g\)
\( For \ { CaCI }_{ 2 },{ CaCI }_{ 2 }\rightleftharpoons { Ca }^{ 2+ }+2{ CI }^{ - } \ i=3\)
\( \triangle { T }_{ f }={ 10 }^{ \circ },{ W }_{ A }=1000g,{ K }_{ f }=1.86 \ K{ m }^{ -1 },{ W }_{ B }=?{ M }_{ B }=111\)
\(10=\frac { 3\times1.86\times{ W }_{ B }\times1000 }{ 1000\times111 } \)
\( or \ { W }_{ B }=\frac { 10\times1000\times111 }{ 3\times1.86\times1000 } =198.92g\)
Therefore, less amount of NaCI is needed and it should be preferred.

Density of water \(=1\ g\ mL^{ -1 }\)
Mass of 1000 mL of water = Volume\(\times \)Density
=1000\(\times \)1 = 1000 g
Moles of water \(\\ =\frac { 1000 }{ 18 } =55.5\)
\(\therefore \) 55.55 moles of \({ H }_{ 2 } O\) are present in 1000 mL or 1 L of water
 Molarity = 55.55 M. 

​​

i) Haloalkanes can easily dissolve in organic solvents of low polarity because the new forces of attraction set up between haloalkanes and the solvent molecules are of same strength as the forces of attraction being broken.
(ii) A mixture of equal amounts of two enantiomers is known as racemic mixture. For example : When a 3° halide undergoes substitution with KOH, the reaction proceeds through S_N1 mechanism forming the racemic mixture in which one of the products has the same configuration as a reactant, while the other product has an inverted configuration.

More stable the carbocation, more reactive will be substrate .the carbocation formed by two compounds are as follows:
Compound (C₆H₅)₂CHBr = carbocation (C₆H₅)₂CH⁺
Compound C₆H₅CH(CH₃)Br = carbocation C₆H₅(CH₃)CH⁺
Out of these two carbocations (C₆H₅)₂CH⁺ is more stable than C₆H₅(CH₃)CH⁺ because the carbocation(C₆H₅)₂CH⁺  is resonance stabilised by two benzene rings therefore (C₆H₅)₂CHBr is more reactive than C₆H₅CH(CH₃)Br. Resonance stabilisation of carbocation with one benzene ring is shown :