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Answer 1 · 2017-07-25T12:31:27

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

So $E \to (100 - 53.33) % = 66.67 %$ $E->(100-53.33)%=66.67%$

So 8 parts of oxygen will combine with $\frac{46.67}{53.33} \cdot 8 \approx 7$ $46.67/53.33*8~~7$ parts of the element $E$ $E$

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

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

We know

$atomic mass = equivalent mass \times valency$ $" atomic mass"="equivalent mass"xx"valency"$

$= 7 v g / m o l$ $=7vg"/"mol$

By the formula the molar mass of the compound will be

$(7 v + 35.5 v) = 42.5 v g / m o l$ $(7v+35.5v)=42.5v g"/"mol$

Given that vapour density of the compound $= 85$ $=85$
So its molar mass $= 2 \times 85 g / m o l$ $=2xx85g"/"mol$

Hence $42.5 v = 2 \times 85$ $42.5v=2xx85$

$\Rightarrow v = 4$ $=>v=4$

So atomic mass of the element

$= 7 v g / m o l = 7 \times 4 = 28 g / m o l$ $=7vg"/"mol=7xx4=28g"/"mol$