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0.532 g of the chloroplatinate of an org...

0.532 g of the chloroplatinate of an organic base (mol.wt 24 D) gave 0.195 g of Pt on ignition. Then the number of nitrogen atoms per molecule of the base is

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
D
`(W)/(2E+410)=(omega)/(195)`
`(0.532)/(2E+410)=(0.195)/(195)rArr E=61`
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