\(\text { (a) } \frac{1}{2}+\frac{3}{2}=2\)
\(\text { (b) } \frac{3}{2}-1=\frac{1}{2}\)

The number of electrons involved in the reaction is three.
Therefore, the amount of electricity required for one mole
of AI³⁺ ions = 3F or 3 x 96500C = 289500C

Suppose B is in excess so that b>> a or x. Neglecting a and x in comparison to b, the equation reduced to k b = k' = \(\frac {2.303}{t} log \frac {a}{(a-x)}\) which is same as for reactions of 1st order.

Normality = 2 × Molarity.

Specific conductance is defined as conductance of electrolyte when distance between electrodes is 1 cm and area of cross section is 1 cm² .

\(\Delta T_{b}\) = 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) The process of osmosis is of great biological and industrial importance as is evident from the following examples: 
(ii) Movement of water from the soil into plant roots and subsequently into the upper portion of the plant occurs partly due to osmosis. 
(iii) Preservation of meat against bacterial action by adding salt. 
(iv) Preservation of fruits against bacterial action by adding sugar. The bacterium in canned fruit loses water through the process of osmosis, shrivels and dies. 
(v) Reverse osmosis is used for the desalination of water. 

(i) Calculate E⁰_cell by using the formula
\(E^{ 0 }_{ cell }=E^{ 0 }_{ cathode }-E^{ 0 }_{ anode }\)
(ii) Calculate the number of electrons by writing the electrode reactions then write Nernst equation and find E_cell
For the cell A1|A1³⁺||Ni²⁺|Ni
(0.001M) (0.50M)
the cell reaction is
\(2AI(s)+3Ni^{ 2+ }(aq)\longrightarrow 2AI^{ 3+ }(aq)+3Ni(s)\)
Thus n=6
\(E^{ 0 }_{ cell }=E^{ 0 }_{ Ni^{ 2+ }/Ni }-E^{ 0 }_{ A13+/A1 }\)
= - 0..25 V- (- 1.66 V) = 1.41 V
\(E_{ cell }=E^{ 0 }_{ cell }=\frac { 0.0591 }{ n } log\frac { \left[ A1^{ 3+ } \right] ^{ 2 } }{ \left[ Ni^{ 2+ } \right] ^{ 3 } } \)
\(E_{ cell }=1.14V-\frac { 0.0591V }{ 6 } log\frac { \left( 10^{ -3 } \right) ^{ 2 } }{ \left( 0.50 \right) ^{ 3 } } \)
\(=1.41V-\frac { 0.0591V }{ 6 } log\frac { 8\times 10^{ 6 } }{ 1 } \left( \therefore \frac { 1 }{ 0.125 } =8 \right) \)
\(E_{ cell }=1.41V-\frac { 0.0591V }{ 6 } \times \left( -5.0969 \right) \)
= 1.41 V + 0..0.50.2 V= 1.460.2 V

The constancy in the value of k proves that the reaction is of the first order.

Rate = k[A₀][B₀][C₀]²,Overall order=1 + 1 + 2 = 4,k = 1.3\(\times\)10^-4 M^-2 min^-1 d[B]/dt = -1.04\(\times\)10^-2M min^-1,d[L]/dt = 6.24\(\times\)10^-3M min^-1.

2.266 \(\times\) 10^-3 gL^-1

\(\triangle { T }_{ b }=373.128 \ K-373.0 \ K=0.128 \ K\)
\(\\ \triangle { T }_{ b }={ K }_{ b } \ \times \ m\)
\( \Rightarrow m=\frac { 0.128 }{ 0.52 } =0.246 \ mol/kg\)
\( Now,\triangle { T }_{ f }={ K }_{ f } \ \times \ m=1.86\times0.246=0.457\)
Freezing point = \(273-0.457=272.543 \ K\) 

(a) Rate of reaction is defined as the change in concentration of reactants or products per unit time. Factors that affect rate of a reaction
(i) Concentration of reactant
(ii) Temperature
\(log{k_2\over k_1}={E_a\over 2.303R}\left[T_2-T_1\over T_1T_2\right]\)
k₁ = 4 x 10^-2 k²
T₁= 8 x 10^-2,
T₁ = 273 + 27 = 300K
T₂ = 273 + 37 = 310K,
R = 8.314 JK^-1mol^-1
\(log{2\times10^{-2}\over 4\times10^{-2}}={E_a\over 2.303\times8.314}\left[310-300\over 300\times310\right]\)
\(log2={E_a\over 2.303\times8.314}\times{10\over 300\times310}\)
\(0.301={E_a\over 2.303\times8.314}\times{10\over 300\times310}\)
\(E_a={0.301\times2.3038.314\times300\times 310\over 10}\)
= 53598 J mol^-1
= 53.598 kJ mol^-1

(ii)The major difference between a primary cell and the secondary cell is that primary cells are the ones that cannot be charged but secondary cells are the ones that are rechargeable.
(iii) Preventions of corrosion :
(1) Barrier protection :
(a) By painting the surface of metal
(b) By coating the surface with thin film of oil or grease.
(c) By electroplating iron with non corrosive metal.
(2) Sacrificial protection : 
In this method iron is coated with a layer of metal more active than iron.
(3) Electrical Protection.
(4) Using anti - rust solutions.

The products of an aqueous solution of NaCl or brine are NaOH, Cl₂, and H₂.
In this reaction, NaCl and H₂O both dissociate as forming chlorine and hydrogen gas and caustic soda.
NaCl ⇾ Na⁺ + Cl^-
H₂O ⇾ H⁺+OH^-
Uses:
1. Sodium hydroxide is used in the manufacturing of soap, paper pulp, and petroleum products.
2. Chlorine gas is used as a disinfectant to kill germs and bacteria in water.
3. Hydrogen gas is used in the production of hydrochloric acid and welding.

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.

(i) It shows positive deviation.
It is due to weaker interaction between acetone and ethanol than ethanol-ethanol interactions.
(ii) Given : W_B  = 10g, W_s = 100 g, W_A = 90 g, M_B = 180 g.mol and d = 1.2 g/mL
\(M=\frac { Wt%\times density\times 10 }{ Mol.wt } \)
\(M=\frac { 10\times 1.2\times 10 }{ 180 } \)
= 0.66 M or 0.66 mol/L
\(m=\frac { { W }_{ B }\times 1000 }{ { M }_{ B }\times { W }_{ A }(in\quad g) } \)
\(m=\frac { 10\times 1000 }{ 180\times 90 } \)
= 0.61 m or 0.61 mol/kg
(or any other suitable method)

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, \({Rate\ of\ 2nd\ reaction\over Rate\ of\ 1st\ reaction}={27R_2\over 8R_1}={27\over 8}\)