A) \[6.25\times {{10}^{-4}}S{{m}^{2}}\,mo{{l}^{-1}}\]
B) \[625\times {{10}^{-4}}S\,{{m}^{2}}\,mo{{l}^{-1}}\]
C) \[62.5\,S\,{{m}^{2}}\,mo{{l}^{-1}}\]
D) \[6250\,S\,{{m}^{2}}\,mo{{l}^{-1}}\]
Correct Answer: A
Solution :
\[k=\frac{1}{R}\times \frac{\ell }{A}\] \[1.3=\frac{1}{50}\times \frac{\ell }{A}\] \[\frac{\ell }{A}=65\,{{m}^{-1}}\] \[{{\Lambda }_{m}}=\frac{K}{100\times molarity}(\frac{K}{1000\times molarity})\] \[=\frac{\left( \frac{1}{2604}\times 64 \right)}{1000\times 0.40}=6.25\times {{10}^{-4}}=\frac{1}{4\times 0.4\times 1000}=\frac{1}{1600}=6.25\times {{10}^{-4}}S\,{{m}^{2}}mo{{l}^{-1}}\]
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