Question

The specific conductance of a saturated solution of AgCl in water is $1.826\times {10}^{-6}{\Omega }^{-1}\text{c}{\text{m}}^{-1}$ at ${25}^{\circ }\text{C}$ calculate its solubility in water at ${25}^{\circ }\text{C}$.
Given that :- ${\lambda }^{\circ }\left(A{g}^{+}\right)=61.92$
${\lambda }^{\circ }\left(\text{C}{\text{l}}^{-}\right)=76.34$

Answer 1

$\kappa =1.826\times {10}^{-6}{\Omega }^{-1}\text{c}{\text{m}}^{-1}$

$\Lambda {}_m^{\circ }=61.92+76.34=138.26$

$\left(aqu\right)=\lambda {}_{+}^{\circ }$ + $\lambda {}_{-}^{\circ }$

$\Lambda {}_m^{\circ }=\frac{\kappa }{c}$ c $\to$ solubility or conc

$c=\frac{\kappa }{\Lambda {}_m^{\circ }}=\frac{1.826\times {10}^{-6}\text{oh}{\text{m}}^{-1}\text{c}{\text{m}}^{-1}}{138.26}$

$=0.01320\times {10}^{-6}$