logo

Ask Questions, Get Answers

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

The resistance of 0.01M of solution of an electrolyte was found to be $150\omega$ at $25^oC$.Calculate equivalent conductance.cell constant =$0.80cm^{-1}$ [For given solution $\eta$ factor =2]

$\begin{array}{1 1}(a)\;265\\(b)\;278\\(c)\;285\\(d)\;255\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
$R=150 \Omega $
$G=\big(\large\frac{1}{150}\big)\Omega ^{-1}$
Cell constant =$0.80cm^{-1}$
K=cell constant $\times$ conductance
$\;\;=0.80\times \large\frac{1}{150}$
$\;\;=5.33\times 10^{-3}\Omega^{-1}cm^{-1}$
Conc.of $C=0.01mol dm^{-3}$
$\lambda_m=\large\frac{K}{C}$
$\Rightarrow \large\frac{5.3\times 10^{-3}}{0.01}$$\times 1000$
$\Rightarrow 530\Omega^{-1}cm^2mol^{-1}$
$\lambda_n-\large\frac{\lambda_m}{\eta}=\frac{530}{2}$
$\Rightarrow 265\Omega^{-1}cm^2eq^{-1}$
Hence (a) is the correct answer.
answered Dec 11, 2013 by sreemathi.v
 

Related questions

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