\({ \pi }_{ cane\ sugar }={ \pi }_{ X }\)
Therefore, c_{cane sugar}=c_X
(where c is molar concentration)
\(\frac { { W }_{ cane\ sugar } }{ { M }_{ cane\ sugar } } =\frac { { W }_{ X } }{ { M }_{ X } } \)
\(\frac { 5g }{ 342 \ g \ { mol }^{ -1 } } =\frac { 0.877 }{ { M }_{ X } } \)
\(\\ \Rightarrow { M }_{ X }=\frac { 0.877\times 342 }{ 5 } g\ { mol }^{ -1 }\)
\(\Rightarrow { M }_{ X }=59.9 \ or \ 60 \ g{ \ { mol }^{ -1 } }\)

Calculation of activation energy of reaction (i)
T₁= 300 K, T₂ = 310 K, k₁ = k, k₂ = 2 k
\(log{k_2\over k_1}={E_A\over 2.303E}\left(T_2-T_2\over T_1T_2\right ),ie.,\ log2={E_a\over 2.303\times8.314}\times{10\over 300\times310}\ or\ E_a=53.60kJmol^{-1}\)
Calculation of rate constant of reaction (i) at 310 K
\(k={0.693\over t_{1/2}}={0.693\over 30\ min}=2.31\times10^{-2}min^{-1}\)
Rate constant of reaction (ii) at 310 K = 2 x 2·31x 10^-2 min^-1 = 4·62 x 10^-2 min^-1
Energy 0f ac tiva tion 0f reac tion (ii) =\({53.60kJ\ mol^{-1}\over 2}=26.80kJ\ mol^{-1}\)
Aim. To calculate k for reaction (ii) at 300 K
\(log{4.62\times10^{-2}\over k_{300}}={26.80\over 2.303\times8.314\times10^{-3}}\times{10\over 300\times310}=0.0151\)
or \({4.62\times10^{-2}\over k_{300k}}=Antilog\ or\ 0151 = 1.035\ or \ k_{300k}={4.62\times10^{-2}\over 1.035}=4.46\times10^{-2}min^{-1}\)

\({ P }_{ 4 }={ 6CI }_{ 2 }\longrightarrow { 4PCI }_{ 3 }\)
\(\left[ { PCI }_{ 3 }+{ 6H }_{ 2 }O\longrightarrow { H }_{ 3 }{ PO }_{ 3 }+3HCI \right] \times 4\)
\({ P }_{ 4 }+{ 6CI }_{ 2 }+{ 12H }_{ 2 }O\longrightarrow { 4H }_{ 3 }{ PO }_{ 3 }+{ 12HCI }\)
1 mole of white phosphorus produces 12 moles of HCI.
\(\frac { 62 }{ 124 }\)mole of white phosphorus produces \(12\times \frac { 62 }{ 124 } =6\)moles of HCI.
Mass of 6 moles of HCI= 6 X 36.5
= 219.0 g of HCI.

(a) \(\Delta { T }_{ f }=i\frac { { K }_{ f }{ W }_{ b }\times 1000 }{ { M }_{ b }\times { W }_{ a } } \)
\(\Delta { T }_{ f }=3\times \left( \frac { 1.86\times 1.9 }{ 95\times 50 } \right) \times 1000\)
= 2.23 K
\(\Delta { T }_{ f }-{ \Delta T }_{ f }^{ ' }=\frac { 273.15-2.23 }{ 273-2.23 } \)
\({ T }_{ f }^{ ' }\) = 270.92 K or 270.77 K
(b) (i) 2 M glucose has a higher boiling point because higher the number of particles lesser is the vapour pressure.
(ii) Reverse osmosis.

(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).

(i) Since compound A on oxidation gives compound 8 with M.F. C₂H₄O₂, therefore, compound 8 may be acetic acid, CH₃COOH and A may be acetaldehyde, CH₃CHO.
(ii) Since compound A, i.e., acetaldehyde reacts with dil. NaOH, therefore, it undergoes aldol condensation to afford an aldol.Further since this aldol on heating gives compound (C), therefore, (C) must be an α, β- unsaturated aldehyde, i.e., but-2-en-I-al (crotonaldehyde).
(iii) Since compound (C) on catalytic hydrogenation gives compound D, therefore, D may be either I-butanal or I-butanol depending upon the extent of hydrogenation.
AIl the reactions involved in this question are explained below:

(i) Since compound (E) with molecular formula, C₃H₆0 does not reduce Tollens' reagent and Fehling's solution but forms 2, 4-dinitrophenylhydrazone, it must be a ketone. But the only possible ketone having the molecular formula, C₃H₆O is acetone or propanone. Thus, compound (E) is acetone or (propanone) CH₃COCH3·
(ii) Since acetone (E) is obtained by heating compound (8) with Ca(OH)₂ therefore, (B) must be acetic acid (ethanoic acid), CH₃COOH.
(iii) Since (D) on oxidation with KMn04 gives acetic acid (8), therefore, (D) must be ethyl alcohol (ethanol), CH₃CH₂OH.
(iv) Since acetic acid (8) and ethyl alcohol (D) are obtained by hydrolysis of (C) under acidic conditions, therefore, (C) must be ethyl acetate (ethyl ethanoate), CH₃COOC₂H₅
(v) Since ethyl acetate (C) and acetic acid (8) are obtained by treatment of compound (A) with ethyl alcohol, therefore, compound (A) must be acetic anhydride (ethanoic anhydride), (CH₃COO)₂O.
(vi) All the reactions involved in this problem can now be explained as follows

(i) Since ketone A (C₄H₈O) undergoes haloforrn reaction, it must contain the grouping CH}CO. Therefore, ketone A (C₄H₈O) must be butanone (CH₃COCH₂CH₃).
(ii) Since ketone A, i.e., butanone gives compound B on reduction, therefore, B must be 2-butanol (CH₃CHOHCH₂CH₃).
(iii) Since B, i.e., 2-butanol on heating with H₂SO₄ gives compound C which forms a mono-ozonide therefore, compound C must be an alkene.
(iv) Since alkene C forms a mono-ozonide D which on hydrolysis in presence of zinc dust (z.e., reductive ozonolysis) gives only acetaldehyde E, therefore, C must be a symmetrical alkene, i.e., 2-butene.
(v) All the reactions involved in the problem can now be explained as follows:

(i) HOCI in presence of an acid generates the reactive electrophile, chloronium ion «r, which attacks benzene to give chlorobenzene (A).
(ii) Chlorobenzene (A) on treatment with NaNH2/liq. NH₃ undergaes-dehydrohalogenation via benzyne to afford aniline (B).
(iii) Aniline (B) on treatment with HBF4 forms the corresponding salt anilinium tetrafluoroborate (C) which on heating with NaNO₂ undergoes Balz-Schiemann reaction through the intermedium formation of benzenediazonium tetrafluoroborate to afford fluorobenzene (D).

If R_I is the rate of first reaction at 25°C, then as its temperature coefficient is 2, its rate at 35°C will be = 2 R₁, at 45°C = 2 x 2 R₁ = 4 R₁ and at 55°C = 2 x 4 R₁ = 8 R₁
If R₂.is the rate of the second reaction at 25°C, then as its temperature coefficient is 3, its rate at 35°C will be = 3 R₂, at 45°C = 3 x 3 R₂ = 9 R₂ and at 55°C = 3 x 9 R₂ = 27 R₂
Also, we are given R₁ = R₂, i.e., at 25°C, the rates are equal.
At 55⁰C, \({Rate\ of\ 2nd\ reaction\over Rate\ of\ 1st\ reaction}={27R_2\over 8R_1}={27\over 8}\)

\(k={0.693\over t_{1/2}}={0.693\over 2.1hr}=0.33hr^{-1}\)
x = 99% of a = 0·99 a
\(t={2.303\over k}log{a\over a-x}={2.303\over 0.33hr^{-1}}log{a\over a-0.99a}={2.303\over 0.33}log 10^2=13.69hours\)
(ii) Amount decomposed = 99% of 6.2 g =\({99\over 100}\times 6.2g=6.138g\)
1 mol NH₂NO₂ (63g) produce N₂O at STP = 22·4 L
6.138 g will produce N₂O at STP =\({22.4\over 63}\times6.138\ L=2.2176L\)

\(2Pb{ { { { (NO }_{ 3 }) } } }_{ 2 }\xrightarrow { heat } 2PbO(s)+{ NO }_{ 2 }+{ O }_{ 2 }\)
                                            Brown (A)
\({ 2NO }_{ 2 }(g)\overset { cooling }{ \rightleftharpoons } { N }_{ 2 }{ O }_{ 4 }(s)\) 
                         (B) Colourless
\({ N }_{ 2 }{ O }_{ 4 }+2NO\overset { heat }{ \underset { 250k }{ \rightleftharpoons } } { 2N }_{ 2 }{ O }_{ 3 }(s)\) 
                             (C) Blue solid

Resonating Structures of  \({ N }_{ 2 }{ O }_{ 4 }\)