A) 2.080%
B) 20.800%
C) 4.008%
D) 40.800%
Correct Answer: C
Solution :
[c] 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 ionization,\[\alpha =\frac{\lambda _{c}^{{}}}{\lambda _{c}^{\infty }}\times 100\] |
\[=\frac{9.54}{238}\times 100=4.008%\] |
You need to login to perform this action.
You will be redirected in
3 sec