Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is $100 Ω$ The conductivity of this solution is 1.29 S $m^{- 1}$ . Resistance of the same cell when filled with 0.2 M of the same solution is 520 $Ω$ . The molar conductivity of 0.02 M solution of the electrolyte will be

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$1240 \times 10^{- 4} S m^{2} {mol}^{- 1}$

$12.4 \times 10 - 4 S m^{2} {mol}^{- 1}$

$124 \times 10^{- 4} S m^{2} {mol}^{- 1}$

$1.24 \times 10^{- 4} S m^{2} {mol}^{- 1}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Given $R = 100 Ω$ and $κ = 1.29 S m^{- 1}$

Cell constant $K_{cell} = κ R = (1.29 S m^{- 1}) (100 Ω) = 129 m^{- 1}$

For 0.02 M solution, we have $κ = \frac{K}{R} = \frac{129 m^{- 1}}{520 Ω}$ and $Λ = \frac{κ}{c} = \frac{(129 / 520) S m^{- 1}}{0.02 \times 10^{3} mol m^{- 3}} = 124 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE