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Question 44 of 52

Q.(i) State the Kohlrausch's law of independent migration of ions.

(ii) The specific conductance or conductivity of a 0.01 M acetic acid at 298 K is 1.65x10-4 S cm-1. Calculate molar conductivity of the solution and degree of dissociation of CH3COOH. Given that, lambda-degree(H+) = 349.1 and lambda-degree(CH3COO-) = 40.9 S cm2 mol-1. (1+2) OR
(i) Write two functions of salt bridge.
(ii) Write Nernst equation of the following galvanic cell and calculate emf of the cell at 298 K temperature: Cu(s) | Cu2+(0.130M) || Ag+(1.0x10-4M) | Ag(s). Given that E-degree(Cu2+/Cu) = +0.34V, E-degree(Ag+/Ag) = +0.80V. (1+2)
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2023Subjective· 3mImportance★★★★★
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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: Λm∘=u+λ+∘+u−λ−∘\Lambda_m^{\circ} = u_+\lambda_+^{\circ} + u_-\lambda_-^{\circ}
  2. Molar conductivity from specific conductance: Λm=κ×1000C=1.65×10−4×10000.01=16.5 S cm2mol−1\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: …

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