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. (a)`1:1`
B. (b)` 1:2`
C. (c )`2:1 `
D. (d)None of these

Answer

B

Explanation

`AgNO_(3)-= 2Ag^(+)+ underset((H_(2)S))(S^(2-)) rarr Ag_(2)S`
`:' 2 "moles "rarr 1" mole" [100xx1=100"millimoles"]`
`:. 100 "millimoles" rarr 50 "millimoles" H_(2)S` required
`CuSO_(4)-= Cu^(+2)+underset((H_(2)S))(S^(2-)) rarr CuS`
`:' 1 "mole" rarr 1"mole" [100xx1=100 "millimoles"]`
`:. 100" millimole" rarr 100 "millimole" H_(2)S` required
Ratio `50/100=1/2`.