A gas mixture of Ne and Ar has a total pressure of 4.00 atm and contains 16.0 mol of gas. If the partial pressure of Ne is 2.75 atm, how many moles of Ar are in the mixture?
A)11.0
B)5.00
C)6.75
D)9.25
E)12.0

Question

Quizplus · Accepted Answer

The partial pressure of Ar can be found by subtracting the partial pressure of Ne from the total pressure of the mixture:

partial pressure of Ar = total pressure - partial pressure of Ne 
partial pressure of Ar = 4.00 atm - 2.75 atm 
partial pressure of Ar = 1.25 atm

Using the ideal gas law, we can determine the number of moles of Ar present:

PV = nRT

n = PV/RT

n = (1.25 atm)(V) / (0.0821 L·atm/mol·K)(T)

We are not given the volume or temperature of the gas mixture, so we cannot solve for n directly. However, we can use the fact that the total number of moles in the mixture is 16.0 mol:

nTotal = nNe + nAr

Solving for nAr:

nAr = nTotal - nNe 
nAr = 16.0 mol - 2.75 mol 
nAr = 13.25 mol

Therefore, the gas mixture contains 13.25 mol of Ar.

A Gas Mixture of Ne and Ar Has a Total