Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
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 $C=0.01mol dm^{-3}$
$\Rightarrow \large\frac{5.3\times 10^{-3}}{0.01}$$\times 1000$
$\Rightarrow 530\Omega^{-1}cm^2mol^{-1}$
$\Rightarrow 265\Omega^{-1}cm^2eq^{-1}$
Hence (a) is the correct answer.
answered Dec 11, 2013 by sreemathi.v

Related questions

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