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NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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Application +4 | -1
The equivalent conductivity of two strong electrolytes at infinite dilution are :
ΛCH3COONa(aq.)∘=91.0Scm2eq.−1{\mathop \mathrm{\Lambda }\limits^ \circ _{\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}\left( {\mathrm{aq}\mathrm{.}} \right)}} = 91.0{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{S {\kern 1pt} {\kern 1pt}c}}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt} \mathrm{eq}{\mathrm{.}^{ - 1}}
ΛHCl(aq.)∘=426.2Scm2eq.−1{\mathop \mathrm{\Lambda }\limits^ \circ _{\mathrm{HCl}\left( {\mathrm{aq}\mathrm{.}} \right)}} = 426.2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{S {\kern 1pt} {\kern 1pt}c}}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt} \mathrm{eq}{\mathrm{.}^{ - 1}}
What additional information one needs to calculate Λ∘\mathop \mathrm{\Lambda }\limits^ \circ ̊ of an aqueous CH3COOH?\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}?
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Correct option: C
Λ∘ofNaCl\mathop \mathrm{\Lambda }\limits^ \circ \mathrm{of}{\kern 1pt} \mathrm{NaCl}
Option C is correct because, according to Kohlrausch's law, to calculate the limiting molar conductivity of CH3COOH, you need the limiting molar conductivities of CH3COONa, HCl, and NaCl.

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