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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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At $25^oC$ the specific conductance of a saturated solution of $AgCl$ is $2.68\times 10^{-4}Sm^{-1}$ and that of water with which the solution was made is $0.86\times 10^{-4}Sm^{-1}$.If molar conductances at infinite dilution of $AgNO_3,HNO_3$ and $HCl$ are respectively,$1.33$$\times 10^{-4},421\times 10^{-4}$ and $426\times10^{-4}Sm^2mol^{-1}$,calculate solubility of $AgCl$in grams per $dm^3$ in water at given temperature

$\begin{array}{1 1}(a)\;1.9\times 10^{-3}\\(b)\;1.89\times 10^{-3}\\(c)\;1.79\times 10^{-3}\\(d)\;1.6\times 10^{-3}\end{array}$

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1 Answer

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$K_{sol}=K_{AgCl}+K_{water}$
$K_{AgCl}=K_{solution}-K_{water}$
$\qquad=(2.68-0.86)\times 10^{-4}Sm^{-1}$
$AgNO_3+HCl\rightarrow AgCl+HNO_3$
By Kohlrauch's law
$\Lambda^o_mAgCl=\Lambda_m^oAgNO_3+\Lambda_m^oHCl-\Lambda_m^oHNO_3$
$\qquad=(133+426-421)\times 10^{-4}Sm^2mol^{-1}$
$\Lambda_m=\large\frac{K}{C}$ and for saturated solution of sparingly soluable salt $\Lambda_m=\Lambda_m^o$
$C=\large\frac{K}{\Lambda_m^o}$
$\;\;=\large\frac{1.82\times 10^{-4}Sm^{-1}}{138\times 10^{-4}Sm^2mol^{-1}}$
$\;\;=1.32\times 10^{-2}molm^{-3}$
$\;\;=1.32\times 10^{-5}moldm^{-3}$
Solubility of AgCl =$(1.32\times 10^{-5}moldm^{-3}){(143.5gmol^{-1})}$
$\Rightarrow 1.89\times 10^{-3}gdm^{-3}$
Hence (b) is the correct answer.
answered Dec 12, 2013 by sreemathi.v
 

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