The ratio of amounts of `H_(2)S` needed to precipitate all the metal ions from `100 ml` of `1 M AgNO_(3)` and `100 ml` of `1 M CuSO_(4)` will be

Options

A. `1:1`
B. `2:1`
C. `1:2``
D. `2:3`

Answer

C

Explanation

`Ag^(+)` ions in `100 ml =0.1 `mol
`Cu^(2+)` ions in `100 ml =0.1` mol
`2Ag^(+) +H_(2)S rarrAg_(2)S +2H^(+)`
`Cu^(2+)+H_(2)S rarrCuS +2H^(+)`
`0.1 `mol `Ag^(+)` require `H_(2)S=0.05` mol,
0.1 mol `Cu^(2+)` require `H_(2)S=0.1` mol
So ratio of `H_(2)S` required `=0.05:0.1=1:2`