# Question 5dbfe

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 $\frac{46.67}{53.33} \cdot 8 \approx 7$ parts of the element $E$

So the equivalent mass of the element will be $7 \frac{g}{\text{equivant}}$

If its valency in chlorde compound be $v$ then formula of its chloride will be $E C {l}_{v}$

We know

$\text{ atomic mass"="equivalent mass"xx"valency}$

$= 7 v g \text{/} m o l$

By the formula the molar mass of the compound will be

$\left(7 v + 35.5 v\right) = 42.5 v g \text{/} m o l$

Given that vapour density of the compound $= 85$
So its molar mass $= 2 \times 85 g \text{/} m o l$

Hence $42.5 v = 2 \times 85$

$\implies v = 4$

So atomic mass of the element

$= 7 v g \text{/"mol=7xx4=28g"/} m o l$