a) 5.24μL He is present per L of air
Pressure, P = 1.000 atm
Volume, V = [tex]5.24 microL * \frac{1 L}{10^{6} microL } = 5.24 * 10^{-6} L[/tex]
Temperature, T = 298.15 K
Using the ideal gas equation to find out moles of He,
[tex]PV = nRT[/tex]
[tex]1.000 atm (5.24 * 10^{-6} L) = n (0.0821\frac{L.atm}{mol.K} )(298.15 K)[/tex]
number of moles n = [tex]2.14 *10^{-7} mol[/tex]
b) Concentration of Ar in % by volume in air = 0.93
That means 0.93 mL Ar is present in 100 mL of air
Moles of Ar = [tex]\frac{PV}{RT}[/tex]
=[tex]\frac{1 atm(0.93mL*\frac{1L}{1000mL}) }{(0.0821Latm/(mol.K))(298.15 K)}[/tex] =[tex]3.80*10^{-5} mol[/tex]
Molarity of Ar in air = [tex]\frac{3.80*10^{-5}mol }{100mL*\frac{1L}{1000 mL} }[/tex]
= [tex]3.80 *10^{-4} mol/L[/tex]
Similarly, mass % of Kr in air = 0.00011 % by volume
0.00011 mL Kr is present per 100 mL of air
Calculating moles of Kr:
n = [tex]\frac{1 atm(0.00011mL*\frac{1L}{1000mL}) }{(0.0821Latm/(mol.K)(298.15K)} =4.49*10^{-9} mol[/tex]
Molarity of Kr in air= [tex]\frac{4.49 * 10^{-9} mol}{100mL*\frac{1L}{1000mL} } = 4.49 *10^{-8} mol/L[/tex]
Mass % of Xe in air is 0.0000087 % by volume
0.0000087 mL Xe is present per 100 mL air
Calculating moles of Xe:
n = [tex]\frac{1atm(0.0000087mL*\frac{1L}{1000mL}) }{0.08206Latm/(mol.K)(298.15K)}= 3.55*10^{-10}mol[/tex]
Molarity of Xe in air = [tex]\frac{3.55*10^{-10}mol }{100mL *\frac{1L}{1000mL} } =3.55*10^{-9} mol/L[/tex]
