In general, specific conductance of an electrolyte decreases with dilution.So,0.002N solution has least specific conductance.

IF = 96500 C = charge of one mole of e^- charge of 6.022 x 10^-23 electron

Given:
λ_∞CH₃COONa = 91.0 ohm^-1 cm²g eq^-1
Formula:
λ_∞CH₃COONa = λ_∞CH₃COONa + λ_∞HCl - λ_∞NaCl
λ_∞HCl = 426.16 ohm^-1 cm²g eq^-1
λ_∞NaCl = 126.45 ohm^-1 cm²g eq^-1
Solution: ∴ λ_∞CH₃COOH
= 91.0 + 426.16 - (126.45)
= 517.16 - 126.45
 λ_∞CH₃COOH = 390.71 ohm^-1cm² g eq^-1

Kohlraush's law:
(i) At infinite dilution, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions. i.e., the molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.
(ii) For a uni - univalent electrolyte such as NaCl, the Kohlraush's law is expressed as
\(\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaCl}}=\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Na}^{+}}+\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Cl}}^{-}\)
(iii) In general, according to Kohlraush's law, the molar conductivity at infinite dilution for a electrolyte represented by the formula A_x B_y, is given below.
\(\left(\Lambda_{m}^{0}\right)_{A_{x} B_{y}}=x\left(\lambda_{m}^{0}\right)_{A^{y+}}+y\left(\lambda_{m}^{0}\right)_{B^{x-}}\)
b) Calculation of molar conductance at infinite dilution for weak electrolytes experimentally.
(i) However, the same can be calculated using Kohlraush's Law. For example, the molar conductance of CH₃COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH₃COONa.
\(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{0} \) ..........(1)
\(\Lambda_{\mathrm{HCl}}^{0}=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0} \) .........(2)
\(\Lambda_{\mathrm{NaCl}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}\) ...........(3)
(ii) Equation (1) + Equation (2) - Equation (3) gives,
\(\left(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}\right)+\left(\Lambda_{\mathrm{HCl}}^{0}\right)-\left(\Lambda_{\mathrm{NaCl}}^{0}\right)=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}-}^{0} \)
\(=\Lambda_{\mathrm{CH}_{2} \mathrm{COOH}}^{0}\)

(i) Faraday is defined as the quantity of electricity required to liberate one gram equivalent of a substance and it is equal to 96,495 coulombs.
(ii) It is denoted by the symbol F Faraday is the quantity of electricity that contains 1 mole of electrons.
(iii) 1 Faraday = 96,495 coulombs = 1 Mole electron.

(i) Oxidation occurs at anode:

(ii) Reduction occurs at cathode:

(iii) Overall reaction:
\({ Zn }_{ (s) }+{ HgO }_{ (s) }\rightarrow { ZnO }_{ (s) }+Hg(l)\)
(iv) Cell emf: about 1.35V.

(i) Specific resistance is denoted by ρ(rho).
(ii) Specific resistance depends on the nature of the electrolyte.
(iii) Reciprocal is specific resistance \(\frac{1}{ρ}\) is specific conductance or conductivity denoted by K (Kappa).
(iv) ohm metre.

Given: [Cu²⁺] = 0.1M; E⁰_Cu²⁺|Cu = +0.34V
Cell reaction is: \(Cu^{2+}_{(aq)}+2e^{-}\rightarrow Cu_{(s)}\)
\(E_{cell}=E^{0}-\frac{0.0591}{n}\log\frac{[Cu]}{[Cu^{2+}]}\)
\(= 0.34\frac{0.0591}{2}\log\frac{1}{0.1}\)
E_cell = 0.34 - 0.0296 = +0.31V

\(\kappa=\mathrm{C}\left(\frac{l}{\mathrm{~A}}\right)\) (or) k = 1/R.l/A
Where, C - Conductance
\(\frac{l}{A} \Rightarrow\) Cell constant (or)
Sp. Conductance \(\propto\) Cell constant
Sp. Conductance \(\propto\) Concentration
1. Sp. Conductance = concentration x cell constant
i) \(\kappa=0.1 \times 0.5=0.05=5 \times 10^{-2} \mathrm{Scm}^{-1}\)
ii) \(\kappa=0.1 \times 0.25=0.025=2.5 \times 10^{-2} \mathrm{Scm}^{-1}\)
2. So the first case will have greater specific conductance with cell constant 0.05.

Kohlrausch deduced the following empirical relationship between the molar conductance (\({ \Lambda }_{ m }\)) and the concentration of the electrolyte (C).
\(​​{ \Lambda }_{ m }={ \Lambda }_{ m }^{ o }-k\sqrt { C } \)

Unit of x
\(k=\frac { 1 }{ \rho } .\frac { l }{ A } \left( \frac { 1 }{ ohm } .\frac { m }{ { m }^{ 2 } } \right) \)
= ohm^-1 m^-1 = mho m^-1 (or) Sm^-1

k = \(\frac{1}{R}.\frac{1}{a}=\frac{1}{ρ}\)
Where K is specific conductance
R is resistance and \(\frac1a\) is cell constant.
ρ is specific resistance.

l = 1.5 cm = 1.5 × 10^-2m 
A = 4.5 cm² = 4.5 × (10^-4)m²
R = 15Ω
\(\kappa = \frac{1}{R}(\frac{l}{A})\)
\(\kappa = \frac{1}{15\Omega}\times\frac{1.5\times10^{-2}m}{4.5\times10^{-4}m^2}\)
= 2.22 ohm^-1 m^-1
= 2.22 Sm^-1

The oxidation potential of M₁ is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.