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Answer 1 · 2016-10-22T09:24:02

Atomic mass of element A be $=43.6" g/mol"$

Let atomic mass of element A be $=a" g/mol"$
Again atomic mass of Oxygen $=16" g/mol"$

So molar mass of the compound $A_2O_3=(2a+48)" g/mol"$
One molecule of the compound $A_2O_3$ contains 2 atoms .

So the ratio of mass of $A$ and mass of $A_2O_3$ in a molecule
$=(2a)/(2a+48)$

Again it is given that 0.359 grams of A reacts to give 0.559 grams of the compound. So we can write

$(2a)/(2a+48)=0.359/0.559$

$=>a/(a+24)=359/559$

$=>1+24/a=559/359=1.55$

$=>24/a=0.55$

$=>a=24/0.55=43.6$