InterviewSolution
Saved Bookmarks
| 1. |
`10800 C` of electricity passed through the electrolyte deposited `2.977g` of metal with atomic mass `106.4 g mol^(-1)`. The charge on the metal cation isA. `+4`B. `+3`C. `+2`D. `+1` |
|
Answer» Correct Answer - a `M^(n+)+n e^(-) rarr M` Mole of metal deposited `=(2.977)/(106.4)=0.27mol` `1 mol `of `M` deposited = `n `Faraday `=nxx96500C.` `0.027 mol =nxx965--xx0.027` `:. Nxx96500xx0.027=10800C` `:. n=4` |
|