Molecular weight of ammonium chloroplatinate (NH4)2PtCl6
= 2{14 + 4(1)} + 195 + 6(35.5)
= 2(18) + 195 + 213
= 444 g
444 g of ammonium chloroplatinate contains 195 g of Pt
∴ 100 g of ammonium chloroplatinate will contain
= 195\over444 × 100
= 43.91% = 44%
Hence, 44% percentage of platinum is present in ammonium chloroplatinate