InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
When a current of 0.75 A is passed through a CuSO_(4) solution for 25 minutes, 0.369 g of copper is deposited at the cathode. Calculate the atomic mass of copper. |
|
Answer» Solution :`Cu^(+)(aq)+UNDERSET(2" moles")(2e^(-)) to underset(1" MOLE")(Cu(s))` QUANTITY of charge passed `=lxxt=(0.75A)xx(25xx60s)` `=1125 A .S.=1125" C"` `1125" C"` of charge deposit copper=0.369 g `2xx96500" C mol"^(-1)` of charge deposit `=((0.369g))/((1125C))xx(2xx96500 " C mol"^(-1))=63.3" g mol"^(-1)`. |
|