\(\underset { Acetylene }{ { HC }{ = }{ CH } } \overset { dill.{ H }_{ 2 }{ SO }_{ 4 },Hg{ SO }_{ 4 } }{ \underset { (Addtion\ of{ H }_{ 2 }O }{ \longrightarrow } } \underset { Vinyl alcocol\\ (unstable) }{ [{ H }_{ 2 }C=CHOH } \overset { tautomerises }{ \longrightarrow } \underset { Acetaldehyde(A) }{ { CH }_{ 3 }{ - }{ CHO } }\overset { dill.NaOH }{ \underset { (Aldol\quad condensation) }{ \longrightarrow } }\)
\( \underset { \beta -Hydroxybutyradehyde(B)\ (An\ aldol) }{ { CH }_{ 3 }{ - }{ CHO }H-{ CH }_{ 2 }-CHO } \overset { Heat }{ \underset { Dehydration\quad (-{ H }_{ 2 }{ O }) }{ \longrightarrow } } \underset { But-2-en-1-al }{ { CH }_{ 3 }{ - }{ CH }=H-CHO } \) 

Molar mass of C₂H4O₂: 12 × 2 + 1 × 4 + 16 × 2 = 60 g mol^-1
\(\text {Moles of } \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}=\frac{2.5 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}=0.0417 \mathrm{~mol}\)
\(\text {Mass of benzene in } \mathrm{kg}=75 \mathrm{~g} / 1000 \mathrm{~g} \mathrm{~kg}^{-1}=75 \times 10^{-3} \mathrm{~kg}\)
\(\text {Molality of } \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}=\frac{\text { Moles of } \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}}{\text { kg of benzene }}=\frac{0.0417 \mathrm{~mol} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{75 \mathrm{~g}}\)
= 0.556 mol kg^-1

Yes, it is similar to Friedal Crafts reaction.

HCOOH < CICH₂COOH < CCl₃COOH < CF₃COOH.

Henry's Law: It states that the partial vapour pressure of gas in vapour phase(g) is directly proportional to the mole fraction of the gas in the solution.
Applications of Henry's Law:
(i) To minimise the painful effects accompanying the decompression of deep sea divers, oxygen diluted with less soluble helium gas is used as breathing gas.
(ii) To increase the solubility of CO₂ in soft, drinks and soda water, the bottle is sealed under high pressure.

\(\Delta T_{b}\) = K_b x m = 0.512 x 0.1 = 0.0512,
i = \(\frac{\text { observed } \Delta \mathrm{T}_{b}}{\text { normal } \Delta \mathrm{T}_{b}}=\frac{0.1536}{0.0512}=3\) 
Solute undergoes dissociation in water. It is a strong electrolyte.

(i) \({ CH }_{ 3 }-\overset { \begin{matrix} O \\ \parallel \end{matrix} }{ C } -{ CH }_{ 3 }\overset { LIAIH_{ 4 } }{ \underset { Reduction }{ \longrightarrow } } { CH }_{ 3 }-\overset { \begin{matrix} OH \\ | \end{matrix} }{ CH } -{ CH }_{ 3 }\\ Propanone\quad \quad \quad \quad \quad \quad \quad \quad \quad Propan-2-ol\)
(b) 

(c) 

(ii) (a) Pentane -2-one gives yellow ppt of iodoform in Idoform test, while pentane-3 one does not give it. 
\({ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }-\overset { \begin{matrix} O \\ \parallel \end{matrix} }{ C } -{ CH }_{ 3 }+3NaOI\longrightarrow { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }COONa+CHI_{ 3 }\downarrow +2NaOH\)
                                                                                   (yellow ppt)                                                 
(b) Ethanal will give iodoform test, while propanal will not give it.
\({ CH }_{ 3 }CHO+3NaOI\longrightarrow CH{ I }_{ 3 }\downarrow +HCOONa+2NaOH\)
                                        (Yellow ppt)

(b) (i)  Ethanal and Propanal
Iodoform test – Methyl ketones give a positive iodoform test.
Ethanal has one methyl group linked to the carbonyl carbon atom and therefore, responds to this test.
Propanal does not respond to this test.
(ii)  Benzaldehyde and Acetophenone
Tollen’s test – aldehydes give a positive Tollen’s test.
Benzaldehyde is an aldehyde. Therefore, it reduces Tollen’s reagent to give a reddish-brown precipitate.
Acetophenone is a ketone and therefore gives a negative test.
(iii)  Phenol and Benzoic acid
Ferric chloride test – phenols give a violet coloration with this test.
Phenol reacts with neutral ferric chloride to form an iron-phenol complex, which gives a violet coloration.
Whereas, benzoic acid reacts with neutral ferric chloride to give a buff coloured precipitate of ferric benzoate.

(a) Osmosis: When a solution is separated from the solvent by a semi-permeable membrane which allows the passage of solvent molecules but does not allow solute particles to pass through it, there is net flow of solvent molecules from the solvent to the solution which is called osmosis.
Osmotic Pressure: Osmotic pressure may be defined as extra pressure that must be applied to the solution side to prevent the flow of solvent into solution through a semi-permeable membrane. Osmotic pressure is determined at room temperature and has appreciable value which can be easily measured
(b) W_B = 1.25 g, W_A = 90 g,
ΔT_b = 80.31- 80.10 = 0.21⁰C,
M_B = ?, K_b = 2.53⁰C kg rnol^-1
\(\Rightarrow\ M_B={W_B\times 1000\over Δ T_b\times W_A}\times K_b\)
\(\Rightarrow\ M_B={1.25\times1000\over 0.21\times 90}\times2.53\)
\(M_B={3162.5\over 18.9}=167.328g\ mol^{-1}\) 

(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

Molar mass of CH3CH2CHCICOOH
= 122.5 g mo^-1
No. of moles of CH₃CH₂CHCICOOH present
\(=\frac{10}{122 \cdot 5}=8.16 \times 10^{-2} \mathrm{~mole}\)
\(\text { Molality }=\frac{8 \cdot 16 \times 10^{-2}}{250} \times 1000=0.3264 \mathrm{~m}\) 
Let a be the degree of dissociation of CH₃CH₂CHCICOOH, then and C be the initial concentration of CH₃CH₂CHCICOOH
\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCOOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCOO}^{-}+\mathrm{H}^{+}\)
Initially                                       0                           0
At Equilibrium C(1-\(\alpha\))              \(c\alpha\)                    \(c\alpha\)
\(\therefore K_{a}=\frac{C^{2} \alpha^{2}}{C(1-\alpha)}=\frac{C \alpha^{2}}{1-\alpha} \simeq C \alpha^{2}\)
\( \{\alpha \text { being very very small) }\\ \therefore K_{a}=C \alpha^{2} \)
\(\alpha=\sqrt{\frac{K_{a}}{C}}=\sqrt{\frac{1 \cdot 4 \times 10^{-3}}{0 \cdot 3264}}=0.065\)

Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g of water.
Molar mass of C₂H₆O₂ = 12 x 2 + 1 × 6 + 16 × 2 = 62 g mol^-1
\(\text {Moles of } \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}=\frac{20 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}=0.322 \mathrm{~mol}\)
\(\text {Moles of water }=\frac{80 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}=4.444 \mathrm{~mol}\)
\(\mathrm{x}_{\text {glycol }}=\frac{\text { moles of } \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}}{\text { moles of } \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}+\text { moles of } \mathrm{H}_{2} \mathrm{O}}\)
\(=\frac{0.322 \mathrm{~mol}}{0.322 \mathrm{~mol}+4.444 \mathrm{~mol}}=0.068\)
\(\text {Similarly, } x_{\text {water }}=\frac{4.444 \mathrm{~mol}}{0.322 \mathrm{~mol}+4.444 \mathrm{~mol}}=0.932\)
Mole fraction of water can also be calculated as: 1- 0.068 = 0.932

(i) \(2CH_{ 3 }COONa\overset { Electrolysis }{ \longrightarrow } 2CH_{ 3 }COO^{ - }+2Na^{ + }\)

At anode : \(2CH_{ 3 }COO^{ - }\longrightarrow \overset { CH_{ 3 } }{ \underset { CH_{ 3 } }{ | } } +2CO_{ 2 }+2e^{ - }\)
At Cathode : \(2H^++2e^-\rightarrow H_2(g)\)
(ii) \(_{ CH_{ 3 }COO }^{ CH_{ 3 }COO }{ _{ \diagup }^{ \diagdown }{ Ca } }\quad \overset { Dry }{ \underset { distillation }{ \longrightarrow } } \underset { Acetone }{ CH_{ 3 }COCH_{ 3 } } +CaCO_{ 3 }\)
(iii) \(\underset { Sodium\quad benzoate }{ C_{ 6 }H_{ 5 }COONa } +\underset { Soda\quad lime }{ NaOH(CaO) } \overset { \triangle }{ \longrightarrow } \underset { Benzene }{ C_{ 6 }H_{ 6 } } +\underset { sodium\\ carbonate }{ Na_{ 2 }CO_{ 3 } } \) 

\(\pi V\)= nRT = \(\frac{{W}^{B}}{{M}^{B}}\times R \times T\)

\(\Rightarrow \)  \(\pi \times1 = \frac{68.4}{342}\times0.0821 \times 300 K \)

\(\Rightarrow \) \(\pi =\frac{24.63}{5} = 4.92 \ atm\)

Given, M₂ (ethylene glycol) = 62 g mol^-1
W₁ = 5.50 kg = 5500 g. \(\Delta T_f\)= 10K
[0⁰​​C - (- 10° C)= 10° C = 10 K]
and K_f = 1.86 K kg mol^-1
\(\Delta T_f=\frac{K_f W_2 1000}{M_2W_1}\)
\(W_2=\frac{\Delta T_f M_2 W_1}{K_f 1000}\)
\(=\frac{10 K \times 62 g mol{-1}\times 5500 g}{1.86 K kg mol^{-1}\times 1000}\)
W₃ = 1833.33 g = 1.833 kg

\(\text { Using relation, } \pi=i C R T=i \frac{n}{V} R T\)
\(n=\frac{\pi V}{i R T}=\frac{0.75 \times 2 \cdot 5}{2 \cdot 47 \times 0 \cdot 0821 \times 300}=0.0308 \text { mole }\)
\( \text { Molar mass of } \mathrm{CaCl}_{2}=40+2 \times 35 \cdot 5=111 \mathrm{~g} \mathrm{~mol}^{-1} \)
\(\therefore \text { Amount of } \mathrm{CaCl}_{2} \text { dissolved }\)
\(=0.0308 \times 111=3.42 \mathrm{~g}\)

\(Molarity \ (M)=\frac{\text { Mass of solute } / \text { molar mass }}{\text { Volume of solution in litres }} \\\)
M = 0.15 M = 0.15 mol L^-1 ;
Molar mass of solute = 7 x 12 + 6 x 1 x 2 x 16 = 122 g mol^-1;
Volume of solution = 250 mL = 0.25 L.
\(\left(0 \cdot 15 \mathrm{~mol} \mathrm{~L}^{-1}\right)=\frac{\text { Mass of solute }}{\left(122 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0 \cdot 25 \mathrm{~L})}\)
Mass of solute = (0.15 mol L^-1) x (122 g mol^-1) x (0.25 L) = 4.575 g 

\(\frac { { p }^{ o }-{ p }_{ s } }{ { p }^{ o } } ={ x }_{ 2 }\) (mole fraction of solute in solution) - Raoult's law.

On increasing the temperature, volume of the solution increases. Hence Molarity decreases.

(a) Carboxylic acids are higher boiling liquids than aldehydes, ketones and even alcohols of comparable molecular masses. This is due to more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding. Hence, both (A) and (R) are correct and (R) is the correct explanation of (A).

 IUPAC name of is butane-dioic acid. For compounds containing more than one carboxyl group, the ending '-e' of the alkane is retained. The number of carboxyl groups are indicated by adding the multiplicative prefix, di, tri etc., to the term 'oic'. Thus both A and R is correct and R is the correct explanation of A.

(a) Ethanol molecules show hydrogen bonding. On adding acetone, its molecules get in between the host molecule and break some of hydrogen bonds between them. Due to weakening of interaction, the mixture of ethanol and acetone shows the positive deviation from Raoult's law. Hence, both (A) and (R) are correct and (R) is the correct explanation of (A).

(a) Polar solute formed interaction with polar solvent and non-polar solute gives interaction with non-polar solvent. Hence, both (A) and (R) are correct and (R) is the correct explanation of (A).

(i) (d): All the reactions are showing the acidic properties of carboxylic acid. Carboxylic acid forms the sodium salts with all i.e., alkali metals, NaOH and Na₂CO₃ etc. and removes the acidic proton from the carboxylic acid.
(ii) (a): In general, greater the +I effect of the group attached to the carboxyl group, lesser will be the acidic strength and greater the -I effect ofthe group, greater will be acidic strength. As number of halogen atoms and electronegativity of halogen atom increases, acidic strength increases. Thus, correct order of acidic strength is
CF₃COOH> CHCl₂COOH > HCOOH > C₆H₅CH₂COOH > CH₃COOH
(iii) (c) : Stronger -I group attached closer to - COOH makes the acid stronger, i.e., acid has the larger dissociation constant. - Br shows poor (-I) effect and also far away from -COOH group i.e., option (c) has smallest dissociation constant.
(iv) (a): Due to ortho-effect, (I) and (II) are stronger acids than (III) and (IV). Due to two ortho-hydroxyl groups in (I), it is stronger acid than (II). (III) is a stronger acid than (IV) because at m-position, -OH group cannot exert its +R effect but can only exert its -I effect while at p-position, -OH group exerts its strong +R effect. Thus, the correct order of acidity is : I > II > III > IV.