A) \[1\times {{10}^{-4}}\]
B) \[0.1\]
C) \[9\times {{10}^{-4}}\]
D) \[1.1\times {{10}^{-5}}\]
Correct Answer: A
Solution :
\[{{\lambda }_{{{m}^{o}}}}=0.05S{{m}^{2}}mo{{l}^{-1}}\] C= 0.009 m \[{{\lambda }_{m}}=0.05S\,{{m}^{2}}mo{{l}^{-1}}\] \[\alpha =\frac{{{\lambda }_{m}}}{{{\lambda }_{{{m}^{o}}}}}=0.1\] For weak acid, \[{{K}_{a}}=\frac{{{\alpha }^{2}}c}{1-\alpha }=\frac{{{(0.1)}^{2}}\times 0.009}{(1-0.1)}={{10}^{-4}}\]You need to login to perform this action.
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