The Molecular weight of a oxide of an element is $44$ . The Equivalent weight of the element is $14$ . The atomic weight of the element is :

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Solution

The correct option is C 14
Let the formula of oxide $= M_{2} O_{x}$ . Its molecular weight $= 44$ .
$2 \times M + x \times 16 = 44$
$2 (E \times x) + x \times 16 = 44$
$2 (14 \times x) + 16 x = 44$
$∴ x = 1$
Atomic weight $= 14 \times 1 = 14$

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