The molar conductance at infinite dilution of $\mathrm{AgNO}_{3}, \mathrm{NaCl}$ and $\mathrm{NaNO}_

Question

The molar conductance at infinite dilution of $\mathrm{AgNO}_{3}, \mathrm{NaCl}$ and $\mathrm{NaNO}_{3}$ are 116.5, 110.3 and 105.2 mho $\mathrm{cm}^{2} \mathrm{~mol}^{-1}$ respectively. The conductance of AgCl in water is $2.40 	imes 10^{-6} \mathrm{mho} \mathrm{cm}^{-1}$ and of water is $1.16 	imes 10^{-6} \mathrm{mho} \mathrm{cm}^{-1}$. The solubility of AgCl in $\mathrm{g} \mathrm{L}^{-1}$ is:\

Professor PC Thomas Classes · Answer

According to Kohlrausch's law the $\Lambda^{0}{ }_{(\mathrm{AgCl})}$ is $121.6 \mathrm{mho} \mathrm{cm}^{2} \mathrm{~mol}^{-1}$ Given $\kappa_{(	ext {AgCl }+ 	ext { water })}=2.40 	imes 10^{-6} \mathrm{mho} \mathrm{cm}^{-1}$\ $	herefore \kappa_{A g C l}=2.40 	imes 10^{-6}-1.16 	imes 10^{-6}$ $$ =1.24 	imes 10^{-6} \mathrm{mho} \mathrm{~cm}^{-1} $$ Since $\Lambda^{V}=\kappa 	imes \frac{1000}{	ext { molarity }}$\21.6 $=1.24 	imes 10^{-6} 	imes \frac{1000}{	ext { molarity }}$ Molarity $=\frac{1.24 	imes 10^{-3}}{121.6} \mathrm{~mol} \mathrm{~L}^{-1}$\ Solubility $=\frac{1.24 	imes 10^{-3}}{121.6} 	imes 143.5 \mathrm{~g} \mathrm{~L}^{-1}$\ $=1.463 	imes 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}$\