Question

The equivalent conductance of silver nitrate solution at $250^{\circ} C$ for an infinite dilution was found to be $133.3 \,\Omega^{-1} cm ^{2}$ equiv $^{-1}$. The transport number of $Ag ^{+}$ions in very dilute solution of $AgNO _{3}$ is $0.464$. Equivalent conductances of $Ag ^{+} $ and $NO _{3}^{-}$ (in $\Omega^{-1} cm ^{2}$ equiv $^{-1}$ ) at infinite dilution are respectively

• A. 195.2, 133.3
• B. 61.9, 71.4
• C. 71.4, 61.9
• D. 133.3, 195.2

Answer 1

Answer: (B)

$\lambda^{\infty}\left(A g^{+}\right)=\text {transport number of } A g^{+} \times \Lambda^{\infty}\left( AgNO _{3}\right)$
$=0.464 \times 133.3 $
$=61.9\, \Omega^{-1} cm ^{2} \text { equiv }^{-1}$
Ry Kohlrausch's law
$\Lambda_{\left( AgNO _{3}\right)}^{\infty} =\lambda_{\left( Ag ^{+}\right)}^{\infty}+\lambda_{\left( NO _{3}^{-}\right)}^{\infty} $
$\left.\therefore \Lambda_{\left( NO _{3}^{-}\right.}^{\infty}\right) =\Lambda_{\left( AgNO _{3}\right)}^{\infty}-\lambda_{\left( Ag ^{+}\right)}^{\infty} $
$=133.3-61.9 $
$=71.4\, \Omega^{-1} cm ^{2} \text { equiv }^{-1}$