Chemistry · JEE Advanced only

Equivalent Conductivity (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 2 to 3 hours. If you already understand molar conductivity and n-factor well, focus on the conversion formulas and practice 8 to 10 numerical problems involving both concepts together..

Last updated: 26 October 2026

Direct answer

Equivalent conductivity is the conducting power of all the ions produced when one gram equivalent of an electrolyte is dissolved in a solution.

Equivalent conductivity is the conducting power of all the ions produced when one gram equivalent of an electrolyte is dissolved in a solution. It connects the solution's ability to conduct electricity to the amount of charge-carrying species present per equivalent weight. While molar conductivity uses moles, equivalent conductivity uses gram equivalents, making it useful when comparing electrolytes with different valencies.

Official syllabus

What the JEE Advanced syllabus says

“specific, equivalent and molar conductivity”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: molar conductivities and their variation with concentration.

JEE Main focuses only on molar conductivity and its variation with concentration. JEE Advanced explicitly includes equivalent conductivity, which requires you to understand gram equivalents and the n-factor concept for electrolytes.

Key ideas

What you must understand

Equivalent conductivity (Λ_eq) = κ × 1000 / N, where κ is specific conductivity and N is normality
Relationship: Λ_eq = Λ_m / n, where n is the n-factor (total charge on cation or anion per formula unit)
At infinite dilution, Λ°_eq for strong electrolytes can be found using Kohlrausch's law with ionic equivalent conductivities
For weak electrolytes, degree of dissociation α = Λ_eq / Λ°_eq
Units: Λ_eq is in S cm² equiv⁻¹ (or Ω⁻¹ cm² equiv⁻¹)
Equivalent conductivity always increases with dilution because interionic attractions decrease

Formulas

Formulas and key facts

Λ_eq = κ × 1000 / N (κ in S cm⁻¹, N in equiv L⁻¹)
Λ_eq = Λ_m / n (n = n-factor of electrolyte)
Λ°_eq = λ°_cation + λ°_anion (Kohlrausch's law for equivalent conductivity)
N = M × n (Normality = Molarity × n-factor)
For H₂SO₄: n = 2, so Λ_eq = Λ_m / 2
For Al₂(SO₄)₃: n = 6, so Λ_eq = Λ_m / 6
α = Λ_eq / Λ°_eq (degree of dissociation for weak electrolytes)

Avoid

Common mistakes

Confusing normality with molarity when calculating Λ_eq. Remember N = M × n, so a 0.1 M H₂SO₄ solution is 0.2 N.
Forgetting to divide Λ_m by n-factor when converting to Λ_eq. For BaCl₂, n = 2, not 1.
Using wrong units. Specific conductivity κ must be in S cm⁻¹ and concentration in equiv L⁻¹ to get Λ_eq in S cm² equiv⁻¹.
Applying Kohlrausch's law incorrectly by using molar ionic conductivities instead of equivalent ionic conductivities.

Worked example

A JEE Advanced-level example

Question. The molar conductivity of 0.05 M BaCl₂ solution at 25°C is 240 S cm² mol⁻¹. Calculate (a) the equivalent conductivity and (b) the specific conductivity of the solution.

Show solution

Identify n-factor for BaCl₂: Ba²⁺ has charge +2, so n = 2
Calculate equivalent conductivity: Λ_eq = Λ_m / n = 240 / 2 = 120 S cm² equiv⁻¹
Find normality: N = M × n = 0.05 × 2 = 0.1 N (or 0.1 equiv L⁻¹)
Rearrange formula for specific conductivity: κ = Λ_eq × N / 1000
Calculate κ = 120 × 0.1 / 1000 = 0.012 S cm⁻¹ (or 1.2 × 10⁻² S cm⁻¹)

Answer: (a) Equivalent conductivity = 120 S cm² equiv⁻¹, (b) Specific conductivity = 0.012 S cm⁻¹ or 1.2 × 10⁻² S cm⁻¹

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Equivalent Conductivity in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is molar conductivities and their variation with concentration.

How long does Equivalent Conductivity take to prepare for JEE Advanced?

About 2 to 3 hours. If you already understand molar conductivity and n-factor well, focus on the conversion formulas and practice 8 to 10 numerical problems involving both concepts together., if you already know the related JEE Main topic.