Verified Solution • 2025–26 Session moderate

QJEEM19Ch5b06: JEE Main 2019 Chemistry - Class 12 JEEM19Ch5b06 Solution: Which one of the following graphs between molar conductivity (…

Which one of the following graphs between molar conductivity ( Λm{\Lambda _m} ) versus C\sqrt C is correct ?

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/648aeb6a7b3e9ee3a7adefdaa4cf6b5d79d25de366165e1a6a7379b68128c741.webp
  • B https://ik.imagekit.io/novatra/photos/147cfde37396c4f4280197043cbb4bc1865dc5ab6974a374596388cb3091406d.webp
  • C https://ik.imagekit.io/novatra/photos/140afe89b9cfbd5710ac0910a04d1bea25152f086a9d775eee7cc28e3ff82c33.webp
  • D https://ik.imagekit.io/novatra/photos/efae1f39a6074f560b7b54cc7c5eb63c7531ffa6db10e4218bc51bf6ed7c94da.webp

Step-by-Step Verified Solution

The graph is drawn using following equation,

Λm=Λm∞−bc{\Lambda _m} = \Lambda _m^\infty - b\sqrt c

As the size of K+ is higher than the size of Na+, then the hydration radii of aqueous Na+ will be more than the aqueous K+. Therefore the ionic mobility of Na+ will be smaller than K+. Hence the conductance of K+ will be higher.

  ∴\therefore Λm∞\Lambda _m^\infty of K+ > Λm∞\Lambda _m^\infty of Na+

⇒\Rightarrow Λm∞\Lambda _m^\infty (KCl) > Λm∞\Lambda _m^\infty (NaCl)

So, y intercept is more for KCl.

As slope b is constant both the lines will be parallel.

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