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The value of $$\wedge _{m}^{\infty }$$ for $$KCl$$ and $$KNO_{3}$$ are $$149.86$$ and $$154.96\:\Omega ^{-1}cm^{2}mol^{-1}$$ respectively. Also $$\lambda _{Cl^{-}}^{\infty }$$ is $$71.44\:ohm^{-1}cm^{2}mol^{-1}$$. The value of $$\lambda _{NO_{3}^{-}}^{\infty }$$ is :


A
76.54 ohm1cm2mol1
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B
133.08 ohm1cm2mol1
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C
37.7 ohm1cm2mol1
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D
none of the above
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Solution

The correct option is A $$76.54\ ohm^{-1}cm^{2}mol^{-1}$$
According to kohlrausch's law of independent migration of ions, the molar conductivity of an electrolyte 
at infinite dilution is equal to the sum of the contributions of the molar conductivites of its ions.

Hence,
$$ { \Lambda  }_{ KNO_3 }^{ \infty  }\quad -\quad { \Lambda  }_{ KCl }^{ \infty  }\quad =\quad { \Lambda  }_{ K^+ }^{ \infty  }\quad +\quad { \Lambda  }_{ NO_3^- }^{ \infty  }\quad -\quad { \Lambda  }_{ K^+ }^{ \infty  }\quad -\quad { \Lambda  }_{ Cl^- }^{ \infty  }\quad =\quad { \Lambda  }_{ NO_3^- }^{ \infty  }\quad -\quad { \Lambda  }_{ Cl^- } $$
Substitute values in the above equation.
$$154.96\quad -\quad 149.86\quad =\quad { \Lambda  }_{ NO_3^- }^{ \infty  } - { 71.44 }$$
$${ \Lambda  }_{ NO_3^- } = 76.54 \   { ohm }^{ -1 }{ cm }^{ 2 }{ mol }^{ -1 }. $$
Hence, option A is correct.

Chemistry

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