Q. The volume occupied by $16\, g$ of oxygen at STP is

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A

$22.4\, dm ^{3}$

B

$44.8\, dm ^{3}$

C

$11.2\, dm ^{3}$

D

$5.6\, dm ^{3}$

Solution:

One mol of $O _{2}=22.4\, dm ^{3}$
i.e, $32\, g _{2}$ of $O _{2}=22.4\, dm ^{3}$
$\therefore 16\, g$ of $O _{2}=11.2\, dm ^{3}$

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