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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

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Solution :`(W)/(2E+410)=(OMEGA)/(195)`
`(0.532)/(2E+410)=(0.195)/(195)RARR E=61`


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