MCQ
Given that

$\Lambda_{ m }^{\alpha}=133.4\,\left( AgNO _{3}\right) ; \Lambda_{ m }^{\alpha}=149.9( KCl )$

$\Lambda_{ m }^{\alpha}=144.9\, S\, cm ^{2} \,mol ^{-1}\left( KNO _{3}\right)$

the molar conductivity at infinite dilution for $AgCl$ is $.......\, S \,cm ^{2}\, mol ^{-1}$

  • A
    $140$
  • $138$
  • C
    $134$
  • D
    $132$

Answer

Correct option: B.
$138$
b
$\wedge_{ m }^{0}( AgCl )=\wedge_{ m }^{0}\left( AgNO _{3}\right)+\wedge_{ m }^{0}( KCl )-\wedge_{ m }^{0}\left( KNO _{3}\right)$

$=133.4+149.9-144.9$

$=138\, S \,cm ^{2}\, mol ^{-1}$

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