In: Chemistry
Part A
A mixture of He, Ar, and Xe has a total pressure of 2.70 atm . The partial pressure of He is 0.450 atm , and the partial pressure of Ar is 0.450 atm . What is the partial pressure of Xe?
Express your answer to three significant figures and include the appropriate units.
Part 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 to three significant figures and include the appropriate units.
A)
Use Dalton law of partial pressure:
Ptotal = p(He) + p(Ar) + p(Xe)
2.70 atm = 0.450 atm + 0.450 atm + p(Xe)
p(Xe) = 1.80 atm
Answer: 1.80 atm
B)
1st calculate the total mol of gas.
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 of gas = mol of N2 + mol of O2 + mol of He
0.8035 mol = 0.250 mol + 0.250 mol + mol of He
mol of 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