Question

A) a concentration of 5.24 ppm he means 5.24 μl of he per liter of air. using the ideal gas law in find how many moles of he are contained in 5.24 μl at 25.00°c (298.15 k) and 1.000 atm. this number is the molarity of he in the air. (2.11x10-7m) (b) find the molar concentrations of ar, kr, and xe in air at 25°c and 1 atm.

Answer 1

a) 5.24μL He is present per L of air

Pressure, P = 1.000 atm

Volume, V =
`5.24 microL * (1 L)/(10^(6) microL ) = 5.24 * 10^(-6) L`

Temperature, T = 298.15 K

Using the ideal gas equation to find out moles of He,

`PV = nRT`

`1.000 atm (5.24 * 10^(-6) L) = n (0.0821(L.atm)/(mol.K) )(298.15 K)`

number of moles n =
`2.14 *10^(-7) mol`

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 =
`(PV)/(RT)`

=
`(1 atm(0.93mL*(1L)/(1000mL)) )/((0.0821Latm/(mol.K))(298.15 K))` =
`3.80*10^(-5) mol`

Molarity of Ar in air =
`(3.80*10^(-5)mol )/(100mL*(1L)/(1000 mL) )`

=
`3.80 *10^(-4) mol/L`

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 =
`(1 atm(0.00011mL*(1L)/(1000mL)) )/((0.0821Latm/(mol.K)(298.15K)) =4.49*10^(-9) mol`

Molarity of Kr in air=
`(4.49 * 10^(-9) mol)/(100mL*(1L)/(1000mL) )  = 4.49 *10^(-8) mol/L`

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 =
`(1atm(0.0000087mL*(1L)/(1000mL)) )/(0.08206Latm/(mol.K)(298.15K))= 3.55*10^(-10)mol`

Molarity of Xe in air =
`(3.55*10^(-10)mol )/(100mL *(1L)/(1000mL) ) =3.55*10^(-9) mol/L`