Plotting $1 / \land_{m}$ against $c \land_{m}$ for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio P/S is
[ $\land_{m}$ = molar conductivity
$\land_{m}^{0}$ = limiting molar conductivity
c = molar concentration
$K_{a}$ = dissociation constant of HX]

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$K_{a} \land_{m}^{0}$

$1 / (K_{a} \land_{m}^{0})$

$2 K_{a} \land_{m}^{0}$

$K_{a} \land_{m}^{0} / 2$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

$\land_{m} = K_{a} = \frac{{cα}^{2}}{1 - α}$
$K_{a} = \frac{c {(\frac{\land_{m}}{\land_{m}^{0}})}^{2}}{1 - \frac{\land_{m}}{\land_{m}^{0}}}$
$K_{a} - K_{a} . \frac{\land_{m}}{\land_{m}^{0}} = c {(\frac{\land_{m}}{\land_{m}^{0}})}^{2}$

$\frac{1}{\land_{m}} = \frac{c \land_{m}}{K_{a} . {(\land_{m}^{0})}^{2}} + \frac{1}{\land_{m}^{0}}$ $\frac{K_{a}}{\land_{m}} - \frac{K_{a}}{\land_{m}^{0}} = \frac{c \land_{m}}{{(\land_{m}^{0})}^{2}} + \frac{1}{\land_{m}^{0}}$
$S = \frac{1}{K_{a} . {(\land_{m}^{0})}^{2}}$
$P = \frac{1}{\land_{m}^{0}}$
$\frac{P}{S} = \frac{1}{\land_{m}^{0}} \times K_{a} . {(\land_{m}^{0})}^{2}$
$= K_{a} . \land_{m}^{0}$