Question

The formula of the chloroplatinate salt is ( left(B H_{2}right)left[P t C l_{6}right] )
( 12 mathrm{g} ) of chloroplatinate salt produces ( 5 mathrm{g} ) of residue which is platinum. ( 195 mathrm{g} ) of platinum will correspond to ( frac{12}{5} times 195=468 mathrm{g} ) of chloroplatinate salt. The molar mass of ( left(B H_{2}right)left[P t C l_{6}right] ) is 468
( mathrm{g} / mathrm{mol} )
The molar mass of ( left[P t C l_{6}right] ) is ( 409.81 mathrm{g} / mathrm{mol} . ) Hence, the molar mass of the base ( left(B H_{2}right) ) will be ( 468-409.81=58 mathrm{g} / mathrm{mol} )

# (c) ü, in i (a) i, ii, iii 107. What is the molar mass of diacidic. ID 5 (b) iii, ii, i (d) ii, i, iii molar mass of diacidic organic Lewis base (B), if 12 g of its chloroplatinate salt (BH,PtCl) on ignition produced 5 g residue of Pt? (C) 88 (d) None of these (b) 58 (a) 52

Solution