Amixture 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.

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

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:

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

Mass fraction of helium: × 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:

= 64,775 g/mol

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

= M

Where:

P is pressure: 150 kPa

R is gas constant: 8,3145

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 × × =

2,3×10⁻³ g/mL

I hope it helps!

i wish i could but this one is weird give me abc and i can solve it.

b.false that your answer