Question

A mixture of noble gases [helium (MW 4), argon (MW 40), krypton (MW 83.8), and xenon (MW 131.3)] is at a total pressure of 150 kPa, and a temperature of 500 K. The mixture has the following composition in mole fraction: 0.25 helium, 0.25 argon, 0.25 krypton. Determine: (a) The mass fraction of helium. (b) The average molecular weight of the mixture. (c) The total molar concentration. (d) The mass density.

Answer 1

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

Step-by-step explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

• Helium: 0,25 moles ×
  `(4 g)/(1 mol)` = 1 g of Helium
• Argon: 0,25 moles ×
  `(40 g)/(1 mol)` = 10 g of Argon
• Krypton: 0,25 moles ×
  `(83,8 g)/(1 mol)` = 20,95 g of krypton
• Xenon: 0,25 moles ×
  `(131,3 g)/(1 mol)` = 32,825 g of Xenon

Total grams: 1g+10g+20,85g+30,825g= 62,675 g

Mass fraction of helium:
`(1 gHelium)/(62,675 g)` × 100 = 1,6%

The mass fraction of Helium is 1,6%

b) Because the mole fraction of all gases is the same the average molecular weight of the mixture is:

`(4+40+83,8+131,3)/(4)` = 64,775 g/mol

c) The molar concentration is possible to know ussing ideal gas law, thus:

`(P)/(R.T)` = M

Where:

P is pressure: 150 kPa

R is gas constant: 8,3145
`(L.kPa)/(K.mol)`

T is temperature: 500 K

And M is molar concentration. Replacing:

M = 3,6×10⁻² M

d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:

3,6×10⁻²
`(mol)/(L)` ×
`(64,775 g)/(mol)` ×
`(1L)/(1000 mL)` =

2,3×10⁻³ g/mL

I hope it helps!