0.3 gm of platinichloride of an organic diacidic base left 0.09 gm of platinum on ignition. The molecular weight of the organic base is:

Options

A. 120
B. 240
C. 180
D. 60

Answer

B

Explanation

Let B be the original base, then
`2B+H_(2)PtCl_(6)rarrB_(2)H_(2)PtCl_(6)overset(Delta)rarrPt`
Eq. wt. of `B_(2)H_(2)PtCl_(6)=2B+2+195+6xx35.5=2B+410`
`("Weight of chloroplatinate")/("Weight of Pt")=("Eq. wt. of salt")/("Eq. wt. of Pt")`
`(0.3)/(0.09)=(2B+410)/(195),` B(Ew) of base = 120
Molecular weight of base = Ew `xx` Acidity = `120xx2=240`