Q.Calculate the molar conductivity of AgI at zero concentration if the molar conductivities of NaI, AgNO$_3$ and NaNO$_3$ at zero concentration are respectively, 126.9, 133.4 and 121.5 $\Omega^{-1}$ cm$^2$ mol$^{-1}$.

A fuel cell continuously turns a fuel + oxidant into electricity; and by Kohlrausch's law of independent ionic migration the common nitrate ion cancels, so Lambda(KNO3) - Lambda(LiNO3) equals Lambda(KCl) - Lambda(LiCl) = 34.9 ohm^-1 cm^2 mol^-1.

1. Fuel cell A fuel cell is a galvanic (voltaic) cell in which the chemical energy released by the combustion-type reaction of a fuel (such as hydrogen, methane or methanol) with an oxidant (oxygen) is converted directly and continuously into electrical energy, as long as the reactants are fed in from outside. In the H2-O2 fuel cell, at the anode H2 + 2OH- -> 2H2O + 2e- and at the cathode O2 + 2H2O + 4e- -> 4OH-, giving an overall reaction 2H2 + O2 -> 2H2O. It is efficient and pollution-free.
2. Using Kohlrausch's law of independent migration of ions At infinite dilution each ion contributes a fixed molar ionic conductance: Lambda(KNO3) = lambda(K+) + lambda(NO3-) Lambda(LiNO3) = lambda(Li+) + lambda(NO3-) Subtracting, the common lambda(NO3-) cancels: Lambda(KNO3) - Lambda(LiNO3) = lambda(K+) - lambda(Li+) The same difference lambda(K+) - lambda(Li+) is obtained from the chlorides (the common lambda(Cl-) cancels there too): lambda(K+) - lambda(Li+) = Lambda(KCl) - Lambda(LiCl) = 149.9 - 115.0 = 34.9 Therefore Lambda(KNO3) - Lambda(LiNO3) = 34.9 ohm^-1 cm^2 mol^-1. Alternative part (B): A lithium-ion battery is a rechargeable cell in which Li+ ions shuttle between a lithiated transition-metal-oxide cathode (e.g. LiCoO2) and a graphite anode through a non-aqueous Li-salt electrolyte during charge/discharge; it is light and gives a high energy density. (The NH4OH Kd part uses Lambda(NH4OH) = Lambda(NH4Cl) + Lambda(NaOH) - Lambda(NaCl) = 129.8 + 217.2 - 109.05 = 238.0 S cm^2 mol^-1, alpha = 9.52/238.0 = 0.04, and Kd = C.alpha^2/(1-alpha) approx 0.01 x (0.04)^2 / 0.96 approx 1.67 x 10^-5.)

Kohlrausch's law lets ionic contributions to conductivity be added independently; applying it to the given data gives a degree of dissociation of about 4.2% for this dilute acetic acid solution.

1. Kohlrausch's law of independent migration of ions states that at infinite dilution, each ion migrates independently of its co-ion, and the limiting molar conductivity of the electrolyte is the sum of the limiting ionic conductivities of the cation and anion: $\Lambda_m^{\circ} = u_+\lambda_+^{\circ} + u_-\lambda_-^{\circ}$
2. Molar conductivity from specific conductance: $\Lambda_m = \kappa \times \frac{1000}{C} = 1.65\times10^{-4} \times \frac{1000}{0.01} = 16.5\ \text{S cm}^2\text{mol}^{-1}$ Limiting molar conductivity via Kohlrausch's law: $\Lambda_m^{\circ} = \lambda^{\circ}(H^+) + \lambda^{\circ}(CH_3COO^-) = 349.1 + 40.9 = 390\ \text{S cm}^2\text{mol}^{-1}$ Degree of dissociation: $\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}} = \frac{16.5}{390} \approx 0.0423$