Question

The specific conductance of $AgCl$ at 298 K is $1.826\times {10}^{-6}oh{m}^{-1}c{m}^{-1}$ . The ionic conductivities of $A{g}^{+}$ and $C{l}^{-}$ ions are $61.92$ and $76.34$ $oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ respectively. What is the solubility of $AgCl$ in water? (Molar mass of $AgCl$ is $143.5gmo{l}^{-1}$)

• A. $1.1\times {10}^{-2}g{L}^{-1}$
• B. $1.9\times {10}^{-3}g{L}^{-1}$
• C. $1.3\times {10}^{-5}g{L}^{-1}$
• D. $2.1\times {10}^{-6}g{L}^{-1}$

Answer 1

The correct option is B $1.9\times {10}^{-3}g{L}^{-1}$

Given, ${\lambda }_{A{g}^{+}}=61.92oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ and

${\lambda }_{C{l}^{-}}=76.34oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

$\therefore {\lambda }_{total}\left(\text{total ionic conductance}\right)=61.92+76.34=138.26oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Given that specific conductance $K=1.826\times {10}^{-6}oh{m}^{-1}c{m}^{-1}$

$\lambda =\frac{1000\times K}{C}$

$\therefore C=\frac{1000\times 1.826\times {10}^{-6}}{138.26}mol{L}^{-3}$

$=>C=1.32\times {10}^{-5}mol{L}^{-1}$

Mass of $AgCl=143.5gmo{l}^{-1}$

$\therefore$ Solubility $=143.5\times 1.32\times {10}^{-5}=1.89\times {10}^{-3}g{L}^{-1}$