Step I
Calculation of the quantity of charge passed:
We know that,
Watt = Ampere x Volt
Ampere =\(\frac { Watt }{ Volt } =\frac { 100 }{ 110 } \)
Now charge = Current x Time
=\(\left( \frac { 100 }{ 110 } amp \right) \times 4\times 60\times 60\)
= 13091 C
Step II
Calculation of mass of Cadmium deposited:
The cathodic reaction is
Cd²⁺_(aq)+2e^-\(\rightarrow\)Cd_(s)
112.2g 2 x 96500C
2 x 96500 C of charge deposited Cd = 112.2 g
13091 C charge will deposite Cd
\(\frac { 112.2g }{ 2\times 96500C } \times 13091C\)
= 7.61g

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

(i) Dil. NaCN forms a complex with Ag₂S which on reduction with zinc gives silver metal. Impurities remain unaffected and can be filtered off
Ag₂S + 4NaCN ➝ 2Na[Ag(CN)₂] + Na₂S
2Na[Ag(CN)₂] + Zn \(\xrightarrow{Zn}\) Na₂[Zn(CN)₄]+ 2Ag
(ii) Iodine reacts with Titanium to form Titanium iodide which on heating strongly gives pure titanium and iodine back.
Ti(impure) + 2I₂ ➝ TiI₄
TiI₄ \(\xrightarrow{heat}\) Ti (pure) + 2I₂
(iii) Cryolite lowers the melting point of alumina and increases electrical conductivity.
In other words, fused Al₂O3 is bad conductor of electricity. Therefore, cryolite (Na₃AIF₆) is added to purified AI₂O₃ which reduces the melting point of AI₂O₃ mixture around 1140 Kand also increases electrical conductance.

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 \({4.62\times10^{-2}\over k_{300k}}=Antilog\ or\ 0151 = 1.035\ or \ k_{300k}={4.62\times10^{-2}\over 1.035}=4.46\times10^{-2}min^{-1}\)

6.0\(\times\) 10²³

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.

Ore particles float because their surface is wetted by oil,i.e.,not easily  wetted by water

(i) Blood is a colloidal solution of albuminoid substance. The styptic action (blood stopping action) of alum is due to coagulation of blood forming a clot which stops further bleeding.
(ii) Values associated with Meera's decision are:
(a) Friendship
(b) Keen observation
(c) Applying knowledge of chemistry to daily life
(d) Responsibility
(e) Love and care.

\(\Delta { T }_{ f }=\frac { { K }_{ f }{ W }_{ b }\times 1000 }{ { M }_{ b }\times { W }_{ a } } \)
\(0.383=\left( \frac { 3.83\times 2.56 }{ M\times 100 } \right) \times 1000\)
M = 256
S \(\times \) x = 256
32 \(\times \) x = 256
x = 8
(b) (i) If we place blood cells in solution containing 1.2% sodium chloride solution, they shrink.
(ii) It we place blood cells in solution containing 0.4% sodium chloride solution, they swell.

(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).

It is defined as the change in concentration of reactant (or product) in a particular time interval i.e.,
Rate of reaction \(=\frac{\text { Change in centration }}{\text { Timse Taken }}\)

Due to higher pressure exerted by steam than in the open pan.

\(\underset{Pyrolusite\\ (A)\ Blackish\ brown}{MnO_2 }+ 4KOH + O_2\xrightarrow{Fuse}\underset{Potassium\ magnate\\ (B)-Green\ coloured}{2K_2MnO_4}+2H_2O\)
\(\underset B{2 KMnO_4} + H20 + (0)_2 \xrightarrow[medium]{Alkaline}\underset{Potasium\ permanganate\\ (C)-Purple\ coloured}{2KMnO_4}+2KOH\)
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.