Question

A gas mixture contains each of the following gases at the indicated partial pressures: N2, 211 torr ; O2, 141 torr ; and He, 159 torr .

Part B

What mass of each gas is present in a 1.15 −L sample of this mixture at 25.0 ∘C?

Answer 1

T = 25 0C
= 298 k
for N2
pN2 = 211 torr
= 0.278 atm
use,
p*v = n*R*T
0.278*1.15 = n*0.082*298
n = 0.0131 mole
(mass of N2)/ = 0.0131*(molar mass of N2)
molar mass of N2 = 28 g/mol
(mass of N2) = 0.0131*28
= 0.367 g

for O2
pO2 = 141 torr
= 0.186 atm
use,
p*v = n*R*T
0.187*1.15 = n*0.082*298
n = 8.8*10^-3 mole
(mass of O2) = 8.8*10^-3*(molar mass of N2)
molar mass of O2 = 32 g/mol
(mass of O2) = 8.8*10^-3*32
= 0.282 g

pHe = 159 torr
= 0.209 atm
use,
p*v = n*R*T
0.209*1.15 = n*0.082*298
n = 9.84*10^-3 mole
(mass of He) = 9.84*10^-3*(molar mass of N2)
molar mass of He = 4 g/mol
(mass of He) = 9.84*10^-3*4
= 0.0394 g