Question #5dbfe

1 Answer
Jul 25, 2017

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#