The limiting molar conductivities for \( H^+, Na^+, Cl^-, CH_3COO^- \) are 349.6, 50.1, 76.3, \( 40.9\, S\, cm^2\, mol^{-1} \), respectively, in water at 298 K.
Match List-I with List-II
|
List-I Electrolyte |
List-II \( Λ°_m \) |
|
(A) \( HCl \) |
(I) \( 390.5\,S\, cm^2\, mol^{-1} \) |
|
(B) \( NaCl \) |
(II) \( 126.4\,S\, cm^2\, mol^{-1} \) |
|
(C) \( CH_3COONa \) |
(III) \( 425.9\,S\, cm^2\, mol^{-1} \) |
|
(D) \( CH_3COOH \) |
(IV) \( 91.0\,S\, cm^2\, mol^{-1} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
By Kohlrausch’s law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, each multiplied by the number of such ions in one formula unit. For a 1:1 electrolyte:
\( \displaystyle \Lambda_m^\circ = \lambda_{+}^\circ + \lambda_{-}^\circ \)
The values given are \( \lambda^\circ \) (S cm2 mol−1): \( \ce{H+} \) 349.6, \( \ce{Na+} \) 50.1, \( \ce{Cl-} \) 76.3, \( \ce{CH3COO-} \) 40.9.
(A) HCl — This matches (III) 425.9 S cm2 mol−1.
\( \displaystyle \Lambda_m^\circ(\ce{HCl}) = \lambda^\circ(\ce{H+}) + \lambda^\circ(\ce{Cl-}) = 349.6 + 76.3 = 425.9\ \mathrm{S\,cm^{2}\,mol^{-1}} \)
(B) NaCl — This matches (II) 126.4 S cm2 mol−1.
\( \displaystyle \Lambda_m^\circ(\ce{NaCl}) = 50.1 + 76.3 = 126.4\ \mathrm{S\,cm^{2}\,mol^{-1}} \)
(C) CH3COONa — This matches (IV) 91.0 S cm2 mol−1.
\( \displaystyle \Lambda_m^\circ(\ce{CH3COONa}) = 50.1 + 40.9 = 91.0\ \mathrm{S\,cm^{2}\,mol^{-1}} \)
(D) CH3COOH — This matches (I) 390.5 S cm2 mol−1. Acetic acid is a weak electrolyte, but its limiting value is still the sum of the ion values.
\( \displaystyle \Lambda_m^\circ(\ce{CH3COOH}) = 349.6 + 40.9 = 390.5\ \mathrm{S\,cm^{2}\,mol^{-1}} \)
Shortcut: Only \( \ce{HCl} \) and \( \ce{CH3COOH} \) contain the very mobile \( \ce{H+} \) ion, so they take the two large values; \( \ce{Cl-} \) is faster than \( \ce{CH3COO-} \), so HCl gets 425.9. (A)-(III) appears in Options (2) and (4), and (C)-(IV) then leaves only Option (4).