An element has atomic mass $31$ . Mass of $1.12$ L at STP of vapours of this element weighs $6.2$ g. The atomicity of this element is :

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Solution

The correct option is A 4
Given, $1.12 L \equiv 6.2 g$
$22.4 L \equiv \frac{6.2 \times 22.4}{1.12} = 124$ = molar mass
$A t o m i c i t y = \frac{124}{31} = 4$
(Atomicity is the total number of atoms present in one molecule of an element or a substance. )

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