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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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For strong electrolytes find $\Lambda_m^{\infty}$ of $CH_3COOH$

Given :

$\Lambda_m^{\infty}(CH_3COONa)=91.1Scm^2mol^{-1}$

$\Lambda_m^{\infty}(HCl)=426.2Scm^2mol^{-1}$

$\Lambda_m^{\infty}(NaCl)=126.5Scm^2mol^{-1}$

$\begin{array}{1 1}(a)\;360\\(b)\;400.92\\(c)\;390.8\\(d)\;396.2\end{array}$

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1 Answer

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By Kohlrausch law
$\Lambda_m^{\infty}CH_3COOH=\Lambda_m^{\infty}CH_3COONa+\Lambda_m^{\infty}HCl-\Lambda_m^{\infty}NaCl$
$\qquad\qquad\quad\;\;\;=91.1+426.2-126.5$
$\qquad\qquad\quad\;\;\;=390.8Scm^2mol^{-1}$
Hence (c) is the correct answer.
answered Dec 12, 2013 by sreemathi.v
 

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