Respuesta :
Answer: B. [tex]2.9\times 10^{21}[/tex]
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
Given : Volume of fresh air = [tex]6.0\times 10^{-4}m^3[/tex]
volume of oxygen =[tex]\frac{20}100}\times 6.0\times 10^{-4}m^3=1.2\times 10^{-4}m^3[/tex]
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = [tex]1.0\times 10^{5}Pa[/tex] = 0.98 atm
V= Volume of the gas = [tex]1.2\times 10^{-4}m^3=0.12L[/tex] [tex]1m^3=1000L[/tex]
T= Temperature of the gas = 300 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= ?
[tex]n=\frac{PV}{RT}=\frac{0.98atm\times 0.12L}0.0821\times 300}=4.8\times 10^{-3}moles[/tex]
1 mole of oxygen contains = [tex]6.023\times 10^{23}[/tex] molecules
Thus [tex]4.8\times 10^{-3}moles[/tex] of oxygen contain=[tex]\frac{6.023\times 10^{23}}{1}\times 4.8\times 10^{-3}=2.9\times 10^{21}[/tex] molecules of oxygen
Thus there are [tex]2.9\times 10^{21}[/tex] oxygen molecules in each breath
