Question

The equivalent conductance of silver nitrate solution at $250^{\circ }C$ for an infinite dilution was found to be $133.3{\Omega }^{-1}c{m}^2$ equiv ${}^{-1}$. The transport number of $A{g}^{+}$ions in very dilute solution of $AgN{O}_3$ is $0.464$. Equivalent conductances of $A{g}^{+}$ and $N{O}_3^{-}$ (in ${\Omega }^{-1}c{m}^2$ 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)

${\lambda }^{\infty }\left(A{g}^{+}\right)=\text{transport number of }A{g}^{+}\times {\Lambda }^{\infty }\left(AgN{O}_3\right)$
$=0.464\times 133.3$
$=61.9{\Omega }^{-1}c{m}^2{\text{ equiv }}^{-1}$
Ry Kohlrausch's law
${\Lambda }_{\left(AgN{O}_3\right)}^{\infty }={\lambda }_{\left(A{g}^{+}\right)}^{\infty }+{\lambda }_{\left(N{O}_3^{-}\right)}^{\infty }$
$\therefore {\Lambda }_{(N{O}_3^{-}}^{\infty })={\Lambda }_{\left(AgN{O}_3\right)}^{\infty }-{\lambda }_{\left(A{g}^{+}\right)}^{\infty }$
$=133.3-61.9$
$=71.4{\Omega }^{-1}c{m}^2{\text{ equiv }}^{-1}$