Question

7. A) A mixture of He, Ar, and Xe has a total pressure of 2.70 atm . The partial pressure of He is 0.200 atm , and the partial pressure of Ar is 0.250 atm . What is the partial pressure of Xe?

B) A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He. The temperature of the mixture is 0 ∘C , and the total pressure is 1.00 atm . How many grams of helium are present in the gas mixture?

Express your answer numerically in grams.

Answer 1

A)

PTotal = p(He) + p(Ar) + p(Xe)

2.70 atm = 0.200 atm + 0.250 atm + p(Xe)

p(Xe) = 2.25 atm

Answer: 2.25 atm

B)

Given:

P = 1.0 atm

V = 18.0 L

T = 0.0 oC

= (0.0+273) K

= 273 K

find number of moles using:

P * V = n*R*T

1 atm * 18 L = n * 0.08206 atm.L/mol.K * 273 K

n = 0.8035 mol

Now use:

total mol = n(N2) + n(O2) + n(He)

0.8035 = 0.250 + 0.250 + n(He)

n(He) = 0.3035 mol

Molar mass of He = 4.003 g/mol

use:

mass of He,

m = number of mol * molar mass

= 0.3035 mol * 4.003 g/mol

= 1.215 g

Answer: 1.22 g