The molar conductivities of KC!, KNO3 , HCl, $NaOAc$ and NaCl are 149.9, 145.0, 426.2, 91.0 and 126.5 S ${cm}^{2} {mol}^{- 1}$ , respectively. The molar conductivity at infinite dilution of acetic acid 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

$390.7 S {cm}^{2} {mol}^{- 1}$

$217.5 S {cm}^{2} {mol}^{- 1}$

$552.7 S {cm}^{2} {mol}^{- 1}$

$517.2 S {cm}^{2} {mol}^{- 1}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We have

$Λ_{m}^{\infty} (HOAc) = Λ_{m}^{\infty} (NaOAc) + Λ_{m}^{\infty} (HCl) - Λ_{m}^{\infty} (NaCl) = (91.0 + 426.2 - 126.5) S {cm}^{2} {mol}^{- 1} = 390.7 S {cm}^{2} {mol}^{- 1}$

The molar conductivity at infinite dilution of acetic acid will be $390.7 S {cm}^{2} {mol}^{- 1}$

Watch 3-min video & get full concept clarity

courses

No courses found

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!