A. 3.01 x 1020
B. 3.01 x 1027
C. 6.02 x 1019
D. 6.02 x 1027
Correct Answer:
Option C = 6.02 x 1019
How many molecules of oxygen would occupy a volume of 2.24 cm³ at s.t.p? [Molar volume at s.t.p = 22,400 cm³, Avogadro’s number = 6.02 x 10²³]
![How many molecules are there in 14 g of nitrogen at s.t.p.? [N = 14, Avogadro's constant = 6.02 x 10^23 mol-1]](https://erudites.ng/wp-content/plugins/contextual-related-posts/default.png "How many molecules are there in 14 g of nitrogen at s.t.p.? [N = 14, Avogadro's constant = 6.02 x 10^23 mol-1]")
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