Question

The equivalent conductance of silver nitrate solution at $250K$ 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 $AgN{O}_3$ is $0.464$. Equivalent conductances of ${Ag}^{+}$ and $N{O}_3^{-}$ (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)

$61.9,71.4$

${\lambda }^{\infty }\left({Ag}^{+}\right)=$ transport number of ${Ag}^{+}\times A_{\left(AgN{O}_3\right)}^{\infty }$

$=0.464\times 133.3=61.9{\Omega }^{-1}{cm}^2{equiv}^{-1}$

By Kohlrausch's law

${\Lambda }_{\left(AgN{O}_3\right)}^{\infty }={\lambda }_{\left({Ag}^{+}\right)}^{\infty }+{\lambda }_{\left(N{O}_3^{-}\right)}^{\infty }$

$\therefore {\Lambda }_{\left(N{O}_3^{-}\right)}^{\infty }={\Lambda }_{\left(AgN{O}_3\right)}^{\infty }-{\lambda }_{\left({Ag}^{+}\right)}^{\infty }$

$=133.3-61.9=71.4{\Omega }^{-1}{cm}^2{equiv}^{-1}$

So, the correct option is $B$