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JEE Main 202126 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
26 August
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2021 · 26 August · Morning Shift · Chemistry · Electrochemistry

Given below are two statements: Statement I : The limiting molar...

Given below are two statements:
Statement I : The limiting molar conductivity of KClKCl (strong electrolyte) is higher compared to that of CH3COOH\mathrm{CH}_{3} \mathrm{COOH} (weak electrolyte).
Statement II : Molar conductivity decreases with decrease in concentration of electrolyte.

In the light of the above statements, choose the most appropriate answer from the options given below:

Given below are two statements: Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher

Options

  1. AStatement I is false but Statement II is true
  2. BBoth Statement I and Statement II are true
  3. CStatement I is true but Statement II is false
  4. DBoth Statement I and Statement II are false

Correct answer

D. Both Statement I and Statement II are false

Step-by-step solution

lon K+{K}^{+} Cl−{Cl}^{-} CH3COO−{CH}_{3}{COO}^{-} H+{H}^{+}
λ∘/(Scm2mol−1){\lambda}^{∘} / \left(S{cm}^{2}{mol}^{- 1}\right) 73.573 . 5 76.376 . 3 40.940 . 9 349.6349 . 6

λ∘M(CH3COOH)=λ∘M(H+)+λ∘M(CH3COO−){\lambda}^{∘}M\left({CH}_{3}COOH\right) = {\lambda}^{∘}M\left({H}^{+}\right) + {\lambda}^{∘}M\left({CH}_{3}{COO}^{-}\right)

=349.6+40.9=390.5Scm2mol−1= 349.6 + 40.9 = 390.5{Scm}^{2}{mol}^{- 1}

λ∘M(KCl)=λ∘M(K+)+λ∘M(Cl−){\lambda}^{∘}M ( KCl ) = {\lambda}^{∘}M\left({K}^{+}\right) + {\lambda}^{∘}M\left({Cl}^{-}\right)

=73.5+76.3=149.8Scm2mol−1= 73.5 + 76.3 = 149.8{Scm}^{2}{mol}^{- 1}

So, statement-I is false.
For an electrolyte the dilution increases (or) decrease in concentration, the number of ions increases due to increase of degree of dissociation. So, the molar conductance increases. Hence, Statement-II is false.

Given below are two statements: Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher

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