Question

Question #5dbfe

Answer 1

The amount of oxygen in the oxide of an element (let it be symolised as `E`) is `53.33%`

So `E->(100-53.33)%=66.67%`

So 8 parts of oxygen will combine with `46.67/53.33*8~~7` parts of the element `E`

So the equivalent mass of the element will be `7g/"equivant"`

If its valency in chlorde compound be `v` then formula of its chloride will be `ECl_v`

We know

`" atomic mass"="equivalent mass"xx"valency"`

`=7vg"/"mol`

By the formula the molar mass of the compound will be

`(7v+35.5v)=42.5v g"/"mol`

Given that vapour density of the compound `=85`
So its molar mass `=2xx85g"/"mol`

Hence `42.5v=2xx85`

`=>v=4`

So atomic mass of the element

`=7vg"/"mol=7xx4=28g"/"mol`