Question

The specific conductance of $AgCl$ at 298 K is $2.326\times {10}^{-6}oh{m}^{-1}c{m}^{-1}$. The ionic conductivities of $A{g}^{+}$ and $C{l}^{-}$ ions are $71.92$ and $86.34$ $oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ respectively. What is the solubility of $AgCl$ in water? (Molarmass of $AgCl$ is $143.5gmo{l}^{-1}$)

• A. $1.11\times {10}^{-2}g{L}^{-1}$
• B. $2.11\times {10}^{-3}g{L}^{-1}$
• C. $1.3\times {10}^{-4}g{L}^{-1}$
• D. $2.11\times {10}^{-6}g{L}^{-1}$

Answer 1

The correct option is B $2.11\times {10}^{-3}g{L}^{-1}$

Given, ${\lambda }_{A{g}^{+}}=71.92oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ and
${\lambda }_{C{l}^{-}}=86.34oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

$\therefore {\lambda }_{total}\left(\text{total ionic conductance}\right)=71.92+86.34=158.26oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Given that specific conductance $K=2.326\times {10}^{-6}oh{m}^{-1}c{m}^{-1}$
$\lambda =\frac{1000\times K}{C}$

$\therefore C=\frac{1000\times 2.326\times {10}^{-6}}{158.26}mol{L}^{-3}$
$=>C=1.47\times {10}^{-5}mol{L}^{-1}$

Mass of $AgCl=143.5gmo{l}^{-1}$
$\therefore$ Solubility $=143.5\times 1.47\times {10}^{-5}=2.11\times {10}^{-3}g{L}^{-1}$