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.