MCQ
Calculate $\Lambda _{HOAc}^\infty $ using appropriate molar conductances of the electrolytes listed above at infinite dilution in ${H_2}O$ at $25\,^o\, C$

Electrolyte :

$KCl$

$KNO_3$

$HCl$

$NaOAc$

$NaCl$

$\Lambda ^\infty (Scm^2mol^{-1}) $:

$149.9$

$145.0$

$426.2$

$91.0$

$126.5$

  • A
    $517.2$
  • B
    $552.7$
  • $390.7$
  • D
    $217.5$

Answer

Correct option: C.
$390.7$
c
(c)$\Lambda _{HOAC}^\infty = \Lambda _{NaOAC}^\infty + \Lambda _{HCl}^\infty - \Lambda _{NaCl}^\infty $

$ = 91.0 + 426.2 - 126.5 = 390.7$

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