The $I P_{1}, I P_{2}, I P_{3} a n d I P_{4}$ Of an element A are $6.0, 10.0, 16.0, a n d 45.0 e v$ respectively. the molecular weight of the oxide of the element A is:
(x is atomic weight)

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Solution

Since $4^{t h}$ $I . P$ is very very high $,$ so it could be Group $13$ element $,$ so assume it be $A l,$ then $A l$ forms $({A l}_{2} O_{3})$ as oxides $.$ $M o ({A l}_{2} O_{3})$ is $101.9693 g {m o l}^{- 1} .$

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