Molar conductivity for a compound AB is 145.0 Scm2mol-1and for CB is 1...
Given:
Molar conductivity of AB = 145.0 Scm²mol⁻¹
Molar conductivity of CB = 110.1 Scm²mol⁻¹
Limiting molar conductivity of A = 73.5 Scm²mol⁻¹
To find: Limiting molar conductivity of C
Formula used:
Kohlrausch law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum of the molar conductivity of its cation and anion, each multiplied by its own migration constant. That is,
Λ°(AB) = λ⁺(A) + λ⁻(B)
Λ°(CB) = λ⁺(C) + λ⁻(B)
where, λ⁺ and λ⁻ are the molar conductivities of the cation and anion, respectively, and the migration constants are represented by Greek letters.
Calculation:
Using Kohlrausch law,
λ⁺(A) + λ⁻(B) = 73.5 Scm²mol⁻¹
λ⁺(C) + λ⁻(B) = 110.1 Scm²mol⁻¹
Subtracting the first equation from the second equation,
(λ⁺(C) + λ⁻(B)) - (λ⁺(A) + λ⁻(B)) = 110.1 Scm²mol⁻¹ - 73.5 Scm²mol⁻¹
λ⁺(C) - λ⁺(A) = 36.6 Scm²mol⁻¹
Therefore, the limiting molar conductivity of C can be calculated as:
λ⁺(C) = λ⁺(A) + 36.6 Scm²mol⁻¹
λ⁺(C) = 73.5 Scm²mol⁻¹ + 36.6 Scm²mol⁻¹
λ⁺(C) = 110.1 Scm²mol⁻¹
Hence, the limiting molar conductivity of C is 110.1 Scm²mol⁻¹.
Therefore, the correct option is (B) 38.6 S cm²mol⁻¹.
Molar conductivity for a compound AB is 145.0 Scm2mol-1and for CB is 1...