Question

At ${18}^{o}C$, the mobility of $A{g}^{+}$ ion is 0.00057 cm $se{c}^{-1}$ and that of $N{O}_3^{-}$ is 0.00063 cm $se{c}^{-1}$ at infinite dilution. What is the molar conductivity of $AgN{O}_3$ at infinite dilution?

• A. $105.8Sc{m}^{2}mo{l}^{-1}$
• B. $115.8Sc{m}^{2}mo{l}^{-1}$
• C. $151.8Sc{m}^{2}mo{l}^{-1}$
• D. None of these

Answer 1

Answer: (B)

$115.8Sc{m}^{2}mo{l}^{-1}$

We have, ${\lambda }_{cation}^{\infty }=u_{cation}^{\infty }\times Faraday=0.00057\times 96500$
${\lambda }_{A{g}^{+}}^{\infty }=55.005Sc{m}^{2}mo{l}^{-1}$
Also, ${\lambda }_{anion}^{\infty }=u_{anion}^{\infty }\times Faraday=0.00063\times 96500$
or ${\lambda }_{N{O}_3^{-}}^{\infty }=60.795Sc{m}^{2}mo{l}^{-1}$
Now from Kohlrausch law
$A_{AgN{O}_3}^{\infty }={\lambda }_{A{g}^{+}}^{\infty }+{\lambda }_{N{O}_3^{-}}^{\infty }=55.005+60.795$
$=115.8Sc{m}^{2}mo{l}^{-1}$