Question

0.532 gram of chloroplatinate of an organic base
(mol wt. 244) gave 0.195 gram of platinum on ignition.
The number of nitrogen atoms per molecule of base is​

Answer 1

Answer:

1.3356 / sec . hello please mrk me as

Answer 2

Answer:

4

Explanation:

molecular weight of choloplatinate = 244u

therefore,                                                      

             2B+H2PtCl6--------->B2H2PtCl2-------(on heat)---->Pt

equivalent wt = 2B+2+195+(6*35.5)

                       = 2B + 410

[wt of chloroplatinate/wt of platinum] = [equivalent wt of salt/equivalent wt pt]

                                        0.532/0.195 = 2B+410/195

therefore,

E = 61

therefore number of nitrogen atoms per molecule of base= 244/61=4

pls mark as brainliest