Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
$\qquad=(2.68-0.86)\times 10^{-4}Sm^{-1}$
$AgNO_3+HCl\rightarrow AgCl+HNO_3$
By Kohlrauch's law
$\qquad=(133+426-421)\times 10^{-4}Sm^2mol^{-1}$
$\Lambda_m=\large\frac{K}{C}$ and for saturated solution of sparingly soluable salt $\Lambda_m=\Lambda_m^o$
$\;\;=\large\frac{1.82\times 10^{-4}Sm^{-1}}{138\times 10^{-4}Sm^2mol^{-1}}$
$\;\;=1.32\times 10^{-2}molm^{-3}$
$\;\;=1.32\times 10^{-5}moldm^{-3}$
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App