A) 0.036 and \[1.66\times {{10}^{-5}}\]
B) 0.121 and \[1.78\times {{10}^{-4}}\]
C) 0.213 and \[1.84\times {{10}^{-5}}\]
D) 0.054 and \[1.94\times {{10}^{-4}}\]
Correct Answer: A
Solution :
Given, \[{{\lambda }_{V}}=14oh{{m}^{-1}}e{{q}^{-1}}\] \[{{\lambda }_{\infty }}=391oh{{m}^{-1}}e{{q}^{-1}}\] \[C=0.0128N\] \[\therefore \] \[\alpha =\frac{{{\lambda }_{v}}}{{{\lambda }_{\infty }}}=\frac{14}{391}=0.036\] Now, \[{{K}_{a}}=C{{\alpha }^{2}}\] \[=0.0128\times {{(0.036)}^{2}}\] \[=1.66\times {{10}^{-5}}\]You need to login to perform this action.
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