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Ozone molecules in the stratosphere absorb much of the harmful radiation from the sun. How many ozone molecules are present in 2.00 L of air under the stratospheric ozone conditions of 275 K temperature and 1.89 × 10−3 atm pressure?

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Answer:

1.01×10^20 molecules of ozone.

Step-by-step explanation:

Data obtained from the question include:

Volume (V) = 2 L

Temperature (T) = 275 K

Pressure (P) = 1.89×10¯³ atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) of ozone =.?

Using the ideal gas equation, we can obtain the number of mole of ozone as follow:

PV = nRT

1.89×10¯³ x 2 = n x 0.0821 x 275

Divide both side by 0.0821 x 275

n = (1.89×10¯³ x 2) /(0.0821 x 275)

n = 1.67×10¯⁴ mole.

Therefore the number of mole of ozone in 2 L of air is 1.67×10¯⁴ mole.

Finally, we shall determine the number of molecules present in 1.67×10¯⁴ mole of ozone.

This can be obtained as follow:

From Avogadro's hypothesis, 1 mole of any substance contains 6.02×10²³ molecules. This implies that 1 mole of ozone contains 6.02×10²³ molecules.

If 1 mole of ozone contains 6.02×10²³ molecules,

therefore, 1.67×10¯⁴ mole of ozone will contain = 1.67×10¯⁴ x 6.02×10²³ = 1.01×10^20 molecules.

Therefore, 1.01×10^20 molecules of ozone are present in 2 L of air.

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