Conductivity and molar conductivity: Class 12 practice

What you should know first

• Units

The idea behind conductivity and molar conductivity

Conductivity κ = cell constant ÷ resistance. Molar conductivity Λm = κ × 1000/C (S cm² mol⁻¹, with κ in S cm⁻¹ and C in mol/L). Kohlrausch's law: Λ°m = λ°₊ + λ°₋.

Step-by-step method

1. κ = G* / R.
2. Λm = 1000κ/C.
3. Λ°m = sum of limiting ionic conductivities.

Worked example 1 (Basic)

A cell of constant 1.3 cm⁻¹ has resistance 1067 Ω with a solution in it. Find the conductivity.

1. κ = 1.3/1067
2. = 1.218e-3 S cm⁻¹
3. Answer: 1.218e-3 S cm⁻¹

Worked example 2 (Standard)

For that solution (κ = 1.083e-3 S cm⁻¹, C = 0.1 M), find the molar conductivity.

1. Λm = 1000κ/C
2. = 10.83 S cm² mol⁻¹
3. Answer: 10.83 S cm² mol⁻¹

Worked example 3 (Challenge)

Limiting ionic conductivities are 73.5 and 78.1 S cm² mol⁻¹. Find Λ°m of the salt (1 : 1).

1. Λ°m = 73.5 + 78.1
2. Answer: 151.6 S cm² mol⁻¹

Common mistakes and misconceptions

• Forgetting the factor 1000
• Inverting the cell constant

Practice questions

1. (Basic) A cell of constant 1.3 cm⁻¹ has resistance 1694 Ω with a solution in it. Find the conductivity.
2. (Basic) A cell of constant 1.2 cm⁻¹ has resistance 341 Ω with a solution in it. Find the conductivity.
3. (Standard) For that solution (κ = 7.519e-4 S cm⁻¹, C = 0.02 M), find the molar conductivity.
4. (Standard) For that solution (κ = 2.215e-3 S cm⁻¹, C = 0.02 M), find the molar conductivity.
5. (Challenge) Limiting ionic conductivities are 119 and 50.1 S cm² mol⁻¹. Find Λ°m of the salt (1 : 1).
6. (Challenge) Limiting ionic conductivities are 50.1 and 50.1 S cm² mol⁻¹. Find Λ°m of the salt (1 : 1).

Answer key with full working

• Q1: κ = 1.3/1694 → = 7.674e-4 S cm⁻¹ → Answer: 7.674e-4 S cm⁻¹
• Q2: κ = 1.2/341 → = 3.519e-3 S cm⁻¹ → Answer: 3.519e-3 S cm⁻¹
• Q3: Λm = 1000κ/C → = 37.59 S cm² mol⁻¹ → Answer: 37.59 S cm² mol⁻¹
• Q4: Λm = 1000κ/C → = 110.73 S cm² mol⁻¹ → Answer: 110.73 S cm² mol⁻¹
• Q5: Λ°m = 119 + 50.1 → Answer: 169.1 S cm² mol⁻¹
• Q6: Λ°m = 50.1 + 50.1 → Answer: 100.2 S cm² mol⁻¹

Syllabus note

This skill is listed under "Electrochemistry" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

• Electrochemistry chapter notes (Class 12)
• Cell EMF and the Nernst equation
• Faraday's laws of electrolysis
• Standard cell EMF and ΔG°
• Electrolysis cells in series
• Molality and mole fraction
• Boiling point elevation and freezing point depression
• First-order reactions and half-life
• Arrhenius equation
• All Class 12 practice

Frequently asked questions

Which chapter is conductivity and molar conductivity in?

"Electrochemistry", Class 12 Chemistry (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.