Problem: In the flask diagrammed below, the left bulb contains 3.0L of neon at a pressure of 2.0 atm. The right bulb contains 2.0L of argon at a pressure of 1.0 atm. What is the mole fraction of neon and argon in the final mixture after the valve is opened and a uniform pressure is reached at constant temperature (25°C).(a) Ne 0.33, Ar 0.67(b) Ne 0.25, Ar 0.75(c) Ne 0.67, Ar 0.33(d) Ne 0.92, Ar 0.82(e) Ne 0.75, Ar 0.25

FREE Expert Solution

We are asked to calculate the mole fractions of each of the gases. 


Mole Fraction (X) relates the moles of components and the total moles:

mole fraction (X)=mole componentmole total


Now to get the total moles we have:

ntotal = nNe + nAr


So for the mole fraction of Ne and Ar we have: 

XNe = nNenNe+nArXAr = nArnNe+nAr


83% (329 ratings)
View Complete Written Solution
Problem Details

In the flask diagrammed below, the left bulb contains 3.0L of neon at a pressure of 2.0 atm. The right bulb contains 2.0L of argon at a pressure of 1.0 atm. What is the mole fraction of neon and argon in the final mixture after the valve is opened and a uniform pressure is reached at constant temperature (25°C).


(a) Ne 0.33, Ar 0.67

(b) Ne 0.25, Ar 0.75

(c) Ne 0.67, Ar 0.33

(d) Ne 0.92, Ar 0.82

(e) Ne 0.75, Ar 0.25

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Mole Fraction concept. You can view video lessons to learn Mole Fraction. Or if you need more Mole Fraction practice, you can also practice Mole Fraction practice problems.