Saved Bookmarks
| 1. |
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S `cm^(-1)` . Calculate its molar conductivity. |
|
Answer» Given `k=2.50xx10^(-2),S^(-1)cm^(-1),M=0.20molL^(-1)` `A_(m)=(kxx1000)/(M)=(1000xx2.50xx10^(-2))/(0.20)` `=125.S^(-1)cm^(2)mol^(-1)` |
|