Question

The specific conductance of a saturated solution of $AgCl$ at $25\,{}^{\circ}C$ is $1.821 \times 10^{-5}\, mho\, cm^{-1}$. What is the solubility of $AgCl$ in water (in $g \,L^{-1}$), if limiting molar conductivity of $AgCl$ is $130.26\, mho\, cm^2\, mol^{-1}$ ?

• A. $1.89 \times 10^{-3}\,g\,L^{-1}$
• B. $2.78 \times 10^{-2}\,g\,L^{-1}$
• C. $2.004\times10^{-2}\,g\,L^{-1}$
• D. $1.43 \times 10^{-3}\,g\,L^{-1}$

Answer 1

Answer: (C)

Solubility $=\frac{\kappa\times1000}{\Lambda^{\circ}_{m}}$
$=\frac{1.821\times10^{-5}\times1000}{130.26}$
$=13.97\times10^{-5}\,mol\,L^{-1}$
$=13.97\times10^{-5}\times143.5\quad\left(AgCl=108+35.5=143.5\right)$
$=2.004\times10^{-2}\,g\,L^{-1}$