The Molecular weight of an 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 A 14
Let the formula of an oxide $= M_{2} O_{x}$ whose $M W$ = $44$
$2 \times M + x \times 16 = 44$
$2 (E \times x) + x \times 16 = 44$
$2 (14 \times x) + 16 x = 44$
$∴ x = 1$
Atomic weight $= 14 \times 1 = 14$

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