QJEEM19Ch037c: JEE Main 2019 Chemistry - Class 11 JEEM19Ch037c Solution: The decreasing order of electrical conductivity of the…

0.1 M Formic acid(HCOOH) (A),
0.1 M Acetic acid(CH₃COOH) (B),
0.1 M Benzoic acid(C₆H₅COOH) (C)

Conductivity (K) depends on no of ions present in unit volume of solution. When no of ions increases conductivity also increases.
Also no of ions depends on degree of dissociation( $\alpha$ ).
We know, $\alpha$ = $\sqrt {{{{K_a}} \over C}}$

For all solutions C = 0.1 M
So, $\alpha$ depends on only K _$a$ here.
K _$a$ of HCOOH = 10^-4
K _$a$ of CH₃COOH = 1.8 $\times$ 10^-5
and K _$a$ of C₆H₅COOH = 6 $\times$ 10^-5

$\therefore$ $\alpha$ (HCOOH) > $\alpha$ (C₆H₅COOH) > $\alpha$ (CH₃COOH)

Therefore conductivity order = A > C > B