Question

The molar conductivities at infinite dilution of $AgN{O}_3$ and $NaCl$ are 116.0 and 110.0 $Sc{m}^{2}mo{l}^{-1}$ respectively. The conductivity of $AgCl$ in water is $8.42\times {10}^{-6}Sc{m}^{–1}$ and of water used is $1.16\times {10}^{-6}Sc{m}^{–1}$. The solubility of $AgCl$ is $6.0\times {10}^{-5}mol/L$. Find the molar conductivity at infinite dilution of $NaN{O}_3$ ($inSc{m}^{-1}mo{l}^{-1}$) in the solution.

Answer 1

Conductivity due to $AgCl$ without $H_{2}O$ is,
$8.42\times {10}^{-6}-1.16\times {10}^{-6}=7.26\times {10}^{-6}Sc{m}^{-1}$

${\lambda }_{AgCl}^o=\frac{\kappa \times 1000}{s}$
$=\frac{7.26\times {10}^{-6}\times 1000}{6.0\times {10}^{-5}}=121Sc{m}^{2}mo{l}^{-1}$

${\lambda }_{AgCl}^o={\lambda }_{AgN{O}_3}^o+{\lambda }_{NaCl}^o-{\lambda }_{NaN{O}_3}^o$
$\Rightarrow {\lambda }_{NaN{O}_3}^o=116+110-121=105Sc{m}^{2}mo{l}^{-1}$