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In an experiment,4 g of M_(2)O_(x) oxide was reduced to 2.8 g of the metal. If the atomic mass of the metal is 56gmol^(-1), the number of O atoms oxide is

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Solution :1 mole of `M_(2)O_(x)` on REDUCTION gives 2 g atoms of M
i.e. `(2xx56+16x)g M_(2)O_(x)" gives M"=2xx56g`
`= 112g`
`therefore 4g M_(2)O_(3)" will GIVE"=(112)/(112+16x)xx6g`
`therefore""(112xx4)/(112+16x)=28`
i.e., `448=2.87(112+16x)`
or `112+16x=160"or"x=3.`


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