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)rarr B_2H_2PtCl_(6) overset(Delta)rarrPt`
Eq.wt. of `B_2H_2 "PtCl"_6 = 2B + 2 + 195 +6 xxx 35.5=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`