Question

At ${25}^{o}C$, the specific conductance of a saturated solution of $AgCl$ is $2.68\times {10}^{-4}S{m}^{-1}$ and that of water with which the solution was made is $0.86\times {10}^{-4}S{m}^{-1}$. If molar conductance at infinite dilution of $AgN{O}_3,HN{O}_3$ and $HCl$ are, respectively, $133.0\times {10}^{-4}$, $421\times {10}^{-4}$ and $426.0\times {10}^{-4}S{m}^{2}mo{l}^{-1}$.
Calculate the solubility of $AgCl$ in grams per $d{m}^3$ in water at the given temperature.

• A. $0.5\times {10}^2$
• B. $1.89\times {10}^{-3}$
• C. $8.65\times {10}^{-3}$
• D. $0.453$

Answer 1

The correct option is B $1.89\times {10}^{-3}$

${\kappa }_{solution}={\kappa }_{AgCl}+{\kappa }_{water}$

$\therefore {\kappa }_{AgCl}={\kappa }_{solution}-{\kappa }_{water}=(2.68-0.86)\times {10}^{-4}S{m}^{-1}=1.82\times {10}^{-4}S{m}^{-1}$

Since, $AgCl$ is formed according to the reaction
$AgN{O}_3+HCl\to AgCl+HN{O}_3$
Hence, using Kohlrauch's law,

${\Lambda }_m^{0}AgCl={\Lambda }_m^{0}AgN{O}_3+{\Lambda }_m^{0}HCl-{\Lambda }_m^{0}HN{O}_3=(133.0+426.0-421.0)\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
$=138.0\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Molar conductivity,
${\Lambda }_m=\frac{\kappa }{C}$

For the saturated solution of sparingle soluble salt,
${\Lambda }_m$=${\Lambda }_m^0$
$$\therefore C=\frac{K}{{\Lambda }_m^0}=\frac{1.82\times {10}^{-4}S{m}^{-1}}{138.0\times {10}^{-4}S{m}^{2}mo{l}^{-1}}=1.32\times {10}^{-2}mol{m}^{-3}$$
$C=1.32\times {10}^{-5}mold{m}^{-3}$

$\text{Solubility in}gd{m}^{-3}=\text{Molar mass}\times C$

$\text{Solubility in}gd{m}^{-3}=143.5gmo{l}^{-1}\times 1.32\times {10}^{-5}mold{m}^{-3}$

$\text{Solubility in}gd{m}^{-3}=1.89\times {10}^{-3}gd{m}^{-3}$