Question

The specific conductance of a saturated solution of silver chloride at ${25}^{o}C$ after subtracting the specific conductance of water is $2.28\times {10}^{-4}S{m}^{-1}$. Calculate the solubility of silver chloride in grams per $d{m}^3$ at this temperature.

${\Lambda }_m^0\left(AgCl\right)=138.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
$M\left(AgCl\right)=143.5gmo{l}^{-1}$

• A. $2.36\times {10}^{-3}g{L}^{-1}$
• B. $7.52\times {10}^{-3}g{L}^{-1}$
• C. $9.88\times {10}^{-3}g{L}^{-1}$
• D. $5.65\times {10}^{-3}g{L}^{-1}$

Answer 1

The correct option is A $2.36\times {10}^{-3}g{L}^{-1}$

Solution:

$\text{Specific conductance,}\kappa =2.28\times {10}^{-4}S{m}^{-1}$

Let the solubility of $AgCl$ be $xmol{m}^{-3}$

Given,
${\Lambda }_m^0\left(AgCl\right)=138.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

$\therefore$
${\Lambda }_m^0=\frac{\kappa }{C}$

${\Lambda }_m^0=\frac{2.28\times {10}^{-4}S{m}^{-1}}{x}$

$x=\frac{2.28\times {10}^{-4}S{m}^{-1}}{138.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}}$

$x=1.648\times {10}^{-2}mol{m}^{-3}$

$x=1.648\times {10}^{-5}mol{L}^{-1}$


$\text{Solubility in}g{L}^{-1}=\text{Molar mass}\times x$

$\text{Solubility in}g{L}^{-1}=143.5gmo{l}^{-1}\times 1.648\times {10}^{-5}mol{L}^{-1}$

$\text{Solubility in}g{L}^{-1}=2.365\times {10}^{-3}g{L}^{-1}$