$W_{1} g$ of an element combined with oxygen forming $W_{2} g$ of its oxide. The equivalent weight of the element is?

[\frac{W_{1}}{W_{2}}] \times 8

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[\frac{W_{1}}{W_{2} - W_{1}}] \times 8

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[\frac{W_{2} - W_{1}}{W_{1}}] \times 8

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[\frac{W_{1}}{W_{1} - W_{2}}] \times 8

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Solution

The correct option is A $[\frac{W_{2} - W_{1}}{W_{1}}] \times 8$
Equivalent weightis the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance.
Given,
$W_{1}$ g of an element forms $W_{2}$ g of its oxide.
Mass of the oxygen= $(W_{2} - W_{1})$
Mass of oxygen=8
Hence, equivalent weight of the element is, $[\frac{W_{2} - W_{1}}{W_{1}}] \times 8$
So,option C is correct answer.

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W1g of an element combined with oxygen forming W2g of its oxide. The equivalent weight of the element is?

$W_{1} g$ of an element combined with oxygen forming $W_{2} g$ of its oxide. The equivalent weight of the element is?