A) \[124\times {{10}^{-4}}S\,{{m}^{2}}\text{ }mo{{l}^{-1}}\]
B) \[1240\times {{10}^{-4}}S\,{{m}^{2}}mo{{l}^{-1}}\]
C) \[1.24\times {{10}^{-4}}S\,{{m}^{2}}mo{{l}^{-1}}\]
D) \[12.4\times {{10}^{-4}}S\,{{m}^{2}}mo{{l}^{-1}}\]
Correct Answer: D
Solution :
\[R=100\,\Omega \] \[k=\frac{1}{R}\left( \frac{l}{a} \right)\] \[\frac{l}{a}\] (cell constant)\[=1.29\times 100\text{ }{{m}^{-1}}\] Given, \[R=520\,W;\,C=0.2\,M\] \[\mu \](molar conductivity) = ? \[\mu =k\times V\] (k can be calculated as \[k=\frac{1}{R}\left( \frac{1}{a} \right);\]now cell constant is known) Hence, \[\mu =\frac{1}{520}\times 129\times \frac{1000}{0.2}\times {{10}^{-6}}{{m}^{3}}\] \[=12.4\times {{10}^{-4}}\text{ }S\,{{m}^{2}}mo{{l}^{-1}}\]
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