# Question #b97d6

Oct 22, 2016

Atomic mass of element A be $= 43.6 \text{ g/mol}$

#### Explanation:

Let atomic mass of element A be $= a \text{ g/mol}$
Again atomic mass of Oxygen $= 16 \text{ g/mol}$

So molar mass of the compound ${A}_{2} {O}_{3} = \left(2 a + 48\right) \text{ g/mol}$
One molecule of the compound ${A}_{2} {O}_{3}$ contains 2 atoms .

So the ratio of mass of $A$ and mass of ${A}_{2} {O}_{3}$ in a molecule
$= \frac{2 a}{2 a + 48}$

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

$\frac{2 a}{2 a + 48} = \frac{0.359}{0.559}$

$\implies \frac{a}{a + 24} = \frac{359}{559}$

$\implies 1 + \frac{24}{a} = \frac{559}{359} = 1.55$

$\implies \frac{24}{a} = 0.55$

$\implies a = \frac{24}{0.55} = 43.6$