Question

The molar conductivity of 0.025 mol L⁻¹ methanoic acid is 46.1 S cm² mol⁻¹. Calculate its degree of dissociation and dissociation constant. Given λ°(H⁺) = 349.6 S cm² mol⁻¹ and λ°(HCOO⁻) = 54.6 S cm² mol⁻¹.

The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 x 10⁻⁴ S cm⁻¹. Calculate molar conductivity ($\Lambda_{m}$) of the solution.

Define the following terms : (i) Fuel cell (ii) Limiting molar conductivity $\Lambda_{m}^{\mathrm{o}}$

Define the following terms : (i) Molar conductivity ($\Lambda_{m}$), (ii) Secondary batteries.

Calculate the molar conductivity and degree of dissociation. Conductivity of 2.5 × 10⁻⁴M methanoic acid is 5.25 × 10⁻⁵ Scm⁻¹. Given : $\Lambda^{0}(\mathrm{H}^{+})=349.5\,\mathrm{S}\,\mathrm{cm}^{2}\,\mathrm{moI}^{-1}\,\mathrm{and}\,\Lambda^{0}(\mathrm{HCOO}^{-})$ = 50.5Scm² mol⁻¹

(i) State the law which helps to determine the limiting molar conductivity of weak electrolyte. (ii) Calculate limiting molar conductivity of CaSO₄ (limiting molar conductivity of calcium and sulphate ions are 119.0 and 160.0 Scm² mol⁻¹ respectively)

Calculate emf of the following cell Cd/\(Cd^{2+}\) (.10 M)//\(H_+\) (.20 M)/\(H_2\) (0.5 atm)/Pt [Given E° for \(Cd^{2+}\) /Cd = -0.403V]