0.984 g of a chloroplatinate salt of a diacid base on ignition gave 0.39 g of platinum. Find the molecular mass of the base.

Explanation

Mass of platinichloride taken `=0.984gm`
Mass of platinum left `=0.39gm`
Step 1. To calculate the equivalent mass of the base.
Let the equivalent mass of the base be B.
Molecular mass of the platininchloride `(B_2H_2PtCL_6)`
`=2B+410`.
Now, `("molecular mass of chloroplatinichloride")/("atomic mass of platinum")`
`=("mass of platinichloride taken")/("mass of platinum left")`
or `(2B+410)/(195)=(0.984)/(0.39)` or `B=(1)/(2)((0.984)/(0.39)xx195-410)=41`
Thus, the equivalent mass of the base be `41`
Step 2: to calculate the molecular mass of the base.
Acidity of the base `=2`
Molecular mass of base
`=`equivalent mass of base`xx`acidity
`=41xx2=82`
Thus, the molecular mass of base is `82`.