Answer to Question #154935 in General Chemistry for Lynil Ruth Caburnay

Question

Given that the quantity of ozone in the stratosphere is equivalent to a 3.0-mm-thick layer of ozone on Earth at STP, calculate the number of ozone molecules in the stratosphere and their mass in kilograms. (Hint: The radius of Earth is 6371 km and the surface area of a sphere is 4 pi r^2, where r is the radius).

Answer 1

The radius of earth (r) = 6371 km

Surface area of a sphere (S.A) = 4πr² = $4× \dfrac{22}7 × (6371 ×10^3)^2$ = 5.1 × 10¹⁴ m²

Volume of the atmosphere volume of atmosphere = S.A × thickness = 5.1 × 10¹⁴ × 3 × 10^-³ = 15.3 × 10¹¹ m³

Since 1m³ = 1000L

15.3 × 10¹¹ m³= 15.3 × 10¹⁴L

Since 22.4 L contains 6.023 × 10²³ Ozone molecules,

Thus, 15.3 × 10¹⁴ L = $\dfrac{6.023×10²³ × 15.3 × 10¹⁴L}{22.4}$ = 4.11 × 10³⁷ molecules of Ozone

1 mole of Ozone (O₃) = 48g (16×3g)

6.023 × 10²³ molecules of Ozone = 48g

4.11 × 10³⁷ molecules of Ozone = $\dfrac{4.11 × 10³⁷ × 48}{6.023× 10²³}$ = 3.275 × 10¹⁵g = 3.275 × 10¹² kg

Therefore, the number of ozone molecules in their stratosphere and their mass in kilograms is 4.11 × 10³⁷ molecules and 3.275 × 10¹² kg respectively.