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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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The dimensions of a conductance cell are 0.15dm & 0.95dm and the electrodes are separated by 0.05dm. What is cell constant for this cell?

$\begin{array}{1 1}(A)\;1/3.45dm^{-1}&(B)\;1/2.85dm^{-1}\\(C)\;1/9.35dm^{-1}&(D)\;1/4.35dm^{-1}\end{array}$

Can you answer this question?
 
 

1 Answer

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Cell constant=$\large\frac{l}{A}$
$\Rightarrow \large\frac{0.05}{0.15\times 0.95}$
$\Rightarrow \large\frac{1}{2.85}$$dm^{-1}$
Hence (B) is the correct answer.
answered Dec 10, 2013 by sreemathi.v
 

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