QJEEM05Ch77a0: JEE Main 2005 Chemistry - Class 12 JEEM05Ch77a0 Solution: .tg {border-collapse:collapse;border-spacing:0;} .tg…

Question

Verified Solution • 2025–26 Session moderate

Electrolyte:	KCl	KNO₃	HCl	NaOAc	NaCl
${ \wedge ^\infty }(Sc{m^2}mo{l^{ - 1}}):$	149.9	145	426.2	91	126.5

Calculate

\wedge _{HOAc}^\infty

Using appropriate molar conductances of the electrolytes listed above at infinite dilution in H₂O at 25^oC

Accepted Answer

A_{HCl}^\infty = 426.2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

A_{AcONa}^\infty = 91.0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

A_{NaCl}^\infty = 126.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)

A_{AcOH}^\infty = \left( i \right) + \left( {ii} \right) - \left( {iii} \right)

= \left[ {426.2 + 91.0 - 126.5} \right]

= 390.7

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