Question

$0.66g$ of platinichloride of a monoacid base left $0.150g$ of platinum on ignition. Calculate its molecular weight. ($Pt=195$)

Answer 1

Let the base is B.

$\therefore B+H_{2}PtC{l}_6\to B_2{H}_{2}PtC{l}_6\to Pt$

Base $0.66g$ $0.150g$

(dibasic}

If W is the weight of the monoacidic base, then molecular wt of plantinchloride

$=2W+2+195+6\times 35.5=(2W+410)g$

Now ($2W+410)g$ of compound contains $195g$ of $Pt$

So $0.66g$ of compound contains $\frac{195}{2W+410}\times 0.66=0.150g$

$\therefore$ Mol. wt of base $=w=224g$