$(a)\;45\qquad(b)\;9\qquad(c)\;18\qquad(d)\;27$

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Let the mass of oxide be 100

$\therefore$ mass of element m is 53

Mass of oxygen is 100-53=47

$\therefore$ Equivalent mass of element =$\large\frac{mass\;of\;element}{mass\;of\;oxygen}$$\times 8$

$\qquad\qquad\qquad\qquad\qquad\;\;=\large\frac{53}{47}$$\times 8$

$\qquad\qquad\qquad\qquad\qquad\;\;=9$(approx)

$\therefore$ Atomic mass=Equivalent mass$\times$ valency

$\Rightarrow 9\times 3$

$\Rightarrow 27amu$

Hence (d) is the correct answer.

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