A mixture of He, Ar, and Xe has a total pressure of 2.30 atm . The partial press

Question

A mixture of He, Ar, and Xe has a total pressure of 2.30 atm . The partial pressure of He is 0.350 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.

Explanation / Answer

A)

Ptotal = sum of all partial pressures

2.30 atm = p(He) + p(Ar) + p(Xe)

2.30 = 0.350 + 0.450 + p(Xe)

p(Xe) = 1.50 atm

Answer: 1.50 atm

B)

1st find the total number of moles,

Given:

P = 1.00 atm

V = 18.0 L

T = 0.0 oC = (0.0+273) K = 273 K

use:

P * V = n*R*T

1.00 atm * 18.0 L = n * 0.0821 atm.L/mol.K * 273.0 K

n = 0.803 mol

This is total number of mol.

use:

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

0.803 = 0.250 + 0.250 + n(He)

n(He) = 0.303 mol

mass of He = mol of He * molar mass of He

= 0.303 mol * 4 g/mol

= 1.212 g

Answer: 1.21 g

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