The molar conductance at infinite dilution of sodium acetate, sodium sulfate and sulfuric acid solutions are
91.0 × 10−4,
259.8 × 10−4 and
859.3 × 10−4 S m2 mol−1, respectively.
Q.24 The molar conductance at infinite dilution (in S m2 mol−1) of acetic acid is
(A) 1028 × 10−4
(B) 820.4 × 10−4
(C) 690.5 × 10−4
(D) 390.8 × 10−4
Introduction
Problems based on molar conductance at infinite dilution are very common in
electrochemistry for JEE, NEET, and other competitive examinations. Such questions require the
application of Kohlrausch’s law of independent migration of ions, especially
for weak electrolytes like acetic acid.
In this article, we calculate the molar conductance at infinite dilution of
acetic acid (CH3COOH) using the given data for strong electrolytes.
🧠 Key Concept Used
According to Kohlrausch’s law, the molar conductance of an electrolyte at
infinite dilution is equal to the sum of the molar ionic conductances of its ions.
This law allows us to calculate the conductance of weak electrolytes using data from strong
electrolytes.
🔢 Step-by-Step Solution
Step 1: Ionic Dissociation
CH3COONa → Na+ + CH3COO−
Na2SO4 → 2Na+ + SO42−
H2SO4 → 2H+ + SO42−
CH3COOH → H+ + CH3COO−
Step 2: Apply Kohlrausch’s Law
Λ∞(CH3COOH) =
Λ∞(CH3COONa)
+ ½ Λ∞(H2SO4)
− ½ Λ∞(Na2SO4)
Step 3: Substitute Values
Λ∞(CH3COOH) =
91.0 + ½(859.3) − ½(259.8)
= 91.0 + 429.65 − 129.9
= 390.75 × 10−4 S m2 mol−1
Step 4: Final Value
Λ∞(CH3COOH) =
390.8 × 10−4 S m2 mol−1
✅ Final Answer
Correct Option: (D)
390.8 × 10−4 S m2 mol−1
📌 Quick Exam Tip
For weak electrolytes, always use Kohlrausch’s law and carefully eliminate
common ions using strong electrolyte data.
🏁 Conclusion
Using Kohlrausch’s law, the molar conductance of acetic acid at infinite dilution can be
accurately calculated by combining conductance values of strong electrolytes. This systematic
approach ensures correct results in electrochemistry numericals.
