0.532 g of the chloroplatinate of a diacid base gave 0.195 g of the platinum residue on ignition. Calaculate the molecular mass of the base.

0.532 g of the chloroplatinate of a diacid base gave 0.195 g of the pl

1.	0.532 g of the chloroplatinate of a diacid base gave 0.195 g of the platinum residue on ignition. Calaculate the molecular mass of the base.
Answer» Solution :Mass of the base chloroplatinate taken `(W) = 0.532 G` Mass of the platinum residue `(x) = 0.195 g` step I. Equivalent mass of the base Equivalent mass of the base `= (1)/(2) [(W)/(x) xx 195 - 410] = (1)/(2) [(0.532)/(0.195) xx 195 - 410]` `= (1)/(2) [532 - 410] = 61` Step II. Molecular mass of the base Molecular mass of the base = Equivalent mass `xx` ACIDITY `= 61 xx 2 = 122`