Question

Molar conductivities (∧°ₘ) at infinite dilution of NaCl, HCl
and CH₃COONa are 126.4, 425.9 and 91.0 S cm² mol⁻¹
respectively. ∧°ₘ for CH₃COOH will be :
(a) 425.5 S cm² mol⁻¹ (b) 180.5 S cm² mol⁻¹
(c) 290.8 S cm² mol⁻¹ (d) 390.5 S cm² mol⁻¹

Answer 1

The molar conductivity is Λ° = 390.5 s cm2 mol-1

Option (D) is correct.

Explanation:

Λ° (HAC) = λ°(H+) + λ°(AC-)  

Λ° = λ° (H+) + λ°(Cl-) + λ° (AC-) + λ° (Na+) – λ°(Cl-) – λ°(Na+)

Λ° = Λ° (HCl) + Λ° (NaAc) – Λ° (NaCl)  

Λ° = (425.9 + 91.0 – 126.4) s cm2 mol-1  

Λ° = 390.5 s cm2 mol-1

Thus the molar conductivity is Λ° = 390.5 s cm2 mol-1

Also learn more

The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm² mol⁻¹. Calculate the conductivity of the solution.  ?

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Answer 2

Answer:

Explanation:

Λ° (HAC) = λ°(H+) + λ°(AC-)  = λ° (H+) + λ°(Cl-) + λ° (AC-) + λ° (Na+) – λ°(Cl-) – λ°(Na+)  = Λ° (HCl) + Λ° (NaAc) – Λ° (NaCl)  = (425.9 + 91.0 – 126.4) s cm2 mol-1  = 390.5 s cm2 mol-1