Question

$0.3$ g of chloroplatinate of an organic diacidic base left $0.09$ g of platinum on ignition. The molecular weight (in g/mol) of the organic base is :

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

Answer 1

The correct option is B $240$

Let $B$ is the original base.
$2B+H_{2}PtC{l}_6\to B_2{H}_{2}PtC{l}_2\stackrel{\Delta }{\to }Pt$

Equivalent weight of $B_2{H}_{2}PtC{l}_2=2B+2+195+6\times 35.5$ $=2B+410$

$\frac{\text{Weight of Chloroplatinate}}{\text{Weight of Pt}}=\frac{\text{Ew of salt}}{\text{Ew of Pt}}$

$⟹\frac{0.3}{0.09}=\frac{2B+410}{195}$

Equivalent weight of base $=120$

Molecular weight of base $=\text{Ew}\times \text{Acidity}$ $=120\times 2=240$ g/mol