A) 2.080%
B) 20.800%
C) 4.008%
D) 40.800%
Correct Answer: C
Solution :
Given, molar conductance at 0.1 M concentration, \[{{\lambda }_{c}}=9.54\,{{\Omega }^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}\] Molar conductance at infinite dilution, \[\lambda _{c}^{\infty }=238\,{{\Omega }^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}\] We know that, degree of ionisation, \[\alpha =\frac{{{\lambda }_{c}}}{\lambda _{c}^{\infty }}\times 100\] \[=\frac{9.54}{238}\times 100=4.008%\]You need to login to perform this action.
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