Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol \(=\frac{\text { moles of ethylene glycol }}{\text { mass of water in kilogram }}\)
\(\text {Moles of ethylene glycol }=\frac{45 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}=0.73 \mathrm{~mol}\)
\(\text {Mass of water in } \mathrm{kg}=\frac{600 \mathrm{~g}}{1000 \mathrm{~g} \mathrm{~kg}^{-1}}=0.6 \mathrm{~kg}\)
\(\text {Hence molality of ethylene glycol }=\frac{0.73 \mathrm{~mol}}{0.60 \mathrm{~kg}}=1.2 \mathrm{~mol} \mathrm{~kg}^{-1}\)
Therefore freezing point depression,
ÄT_f = 1.86 K kg mol^–1 x 1.2 mol kg^–1 = 2.2 K
Freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K

K = 5.25 x 10^-5 S cm^-1, M = 2.5 x 10^-4 M.
\(\lambda_{\mathrm{HCOOH}}^{0}=\lambda_{\left(\mathrm{HCOO}^{-}\right)}^{0}+\lambda_{\mathrm{H}}^{0}+\) 
= 349.5 + 50.5 : 400 S cm² mol^-1.
\(\Lambda_{m}=\frac{1000 \kappa}{\mathrm{M}}=\frac{1000 \times 5.25 \times 10^{-5}}{2.5 \times 10^{-4}}\) 
= \(\frac{1000 \times 525}{10 \times 2.5 \times 100}\) 
\(\Rightarrow \quad \Lambda_{m}=\frac{525}{2.5}=\frac{5250}{25}=210 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) 
\(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{0}}=\frac{210}{400}=\frac{21}{40}=0.525\) 
\(\Rightarrow \ \alpha\) =  0.525 x 100% = 52.5%

BCC

1. Place the egg in dil. HCl or dil. H₂SO₄ solution.
2.  After sometime , outer shell of egg dissolves.
3. Egg with only semi-permeable membrane is now removed and placed in hypertonic solution.
4. After sometime. the size of the egg gets reduced due to osmosis.
5. Egg is now placed in bottle with narrow neck
6. Add hypotonic solution in the bottle containing egg.
7. Egg regains its shape due to osmosis.

(a) \(\Delta_{r} G^{\circ}=-n \mathrm{E}^{\circ} \mathrm{F}\) 
= -2 x 2.71 V x 96500 C
= - 523030 J
= - 523.030 kJ mol^-1

(b) H₂ -O₂ Fuel Cell was used in Apollo space programme for providing electrical power.

Boiling points depend upon the extent of H-bonding which, in turn, depends upon the number of H-atoms present on the N-atom. Since C₂H₉NH₂ has two, (C₂H₅)₂NH has one and C₂H₅N(CH₃)₂ has no hydrogen linked to nitrogen, therefore, boiling points decrease as the extent H-bonding decreases, i.e., boiling points decrease in the order :
C₄H₉NH₂ > (C₂H₅)₂NH > C₂H₅N(CH₃)₂.

\(\underset{Pyrolusite\\ (A)\ Blackish\ brown}{MnO_2 }+ 4KOH + O_2\xrightarrow{Fuse}\underset{Potassium\ magnate\\ (B)-Green\ coloured}{2K_2MnO_4}+2H_2O\)
\(\underset B{2 KMnO_4} + H20 + (0)_2 \xrightarrow[medium]{Alkaline}\underset{Potasium\ permanganate\\ (C)-Purple\ coloured}{2KMnO_4}+2KOH\)
or \(MnO^{-2}_4 \xrightarrow{} MnO_4^- + e^-\)
(b) When acidic solution of green compound (B), i·.e., potassium manganate is allowed to stand for some time, it disproportionates to give permanganate as follows:
\(3 MnO^{2-}_4+ 4H^+\longrightarrow2 MnO^-_4 + MnO_2 + 2 H_2O\)
This reaction is called disproportionation reaction.

Alkaline KMnO4, acidified K₂Cr₂O₇ or HNO₃ cannot be used since all of these will cleave the molecule at the site of the double bond giving a mixture of ketones/acids. The most suitable reagent for this oxidation is NaOI (I₂/NaOH) since methyl ketones on treatment with NaOI undergo iodoform reaction to give iodoform along with the Na salt of a carboxylic acid having one carbon atom less than the starting methyl ketone
        

Higher the oxidation potential, more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be Ag < Hg < Cr < Mg < K.

Hydrated salts like CuSO₄ . 5 H₂O.

NaCl must have been subjected to high pressure.

It is because reaction does not stop in one step.  It can for disubstituted product also.

(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) Chemical formula of ferric chloride is FeCI₃.
(ii) Fe³⁺ ions of FeCl₃ neutralize the charge on the colloidal particles of blood. This leads to coagulation of blood. Bleeding therefore, stopped.
(iii) This phenomenon is known as coagulation or flocculation.
(iv) All house wives must keep small bag or kit in their kitchen. It must have ferric chloride kept in small bottle or potash alum, burnol and bandages etc. Minor accidents are very common in kitchen. The kit can be very helpful to deal with emergency skill of applying knowledge, helping others, kindness, etc are the values associated with it.

(i) India is a tropical country and summer extend over a long period. This means solar power plants are very effective and can be a better substitute of electricity.
(ii) We must hold seminars with the help of NGO's and other organisations. Educate people about the use of solar heaters, solar generators, solar cookers and solar batteries. We can explain the advantages of the use of solar power over electricity.

(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).

\(k={0.693\over t_{1/2}}={0.693\over 2.1hr}=0.33hr^{-1}\)
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 =\({22.4\over 63}\times6.138\ L=2.2176L\)