Question

The molar conductance at infinite dilution of $AgN{O}_3$, $AgCl$ and $NaCl$ are $116.5$, $121.6$ and $110.3$ respectively. The molar conductance of $NaN{O}_3$ is:

• A. $111.4$
• B. $105.2$
• C. $130.6$
• D. $150.2$

Answer 1

Answer: (B)

$105.2$

${\Lambda }_m^{\infty }\left(AgN{O}_3\right)=116.5Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\infty }\left(AgCl\right)=121.6Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\infty }\left(NaCl\right)=110.3Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\infty }\left(AgN{O}_3\right)={\Lambda }^{\infty }A{g}^{+}+{\Lambda }^{\infty }N{O}_3^{-}=116.5$.....(i)

${\Lambda }_m^{\infty }\left(AgCl\right)={\Lambda }^{\infty }A{g}^{+}+{\Lambda }^{\infty }C{l}^{-}=121.6$.....(ii)

${\Lambda }_m^{\infty }\left(NaCl\right)={\Lambda }^{\infty }N{a}^{+}+{\Lambda }^{\infty }C{l}^{-}=110.3$.....(iii)

$\left(i\right)+\left(iii\right)-\left(ii\right)$, we get-

${\Lambda }_m^{\infty }\left(NaN{O}_3\right)={\Lambda }_m^{\infty }AgN{O}_3+{\Lambda }_m^{\infty }NaCl-{\Lambda }_m^{\infty }AgCl$

$={\Lambda }^{\infty }N{a}^{+}+{\Lambda }^{\infty }N{O}_3^{-}$

$=116.5+110.3-121.6$

$=105.2$

The molar conductance of $NaN{O}_3$ is $105.2Sc{m}^{2}mo{l}^{-1}$