Solving a Problem: Calculating Molar Fraction of Freon in a Tank

[simonak]

Hi,

here's the problem :

A 60 gal tank contain pure Azote at 36,2 psig and 24,2 °C. We add 908,25 g of freon (CF2Cl2) that evaporate completely. If the temperature of the tank is 8,2 °C :

a) What is the molar fraction of the freon.


I've made some calculation, but it always result that I need the volume of the Azote to continue.

 

[Gokul43201]

Show us your calculations, and we can tell you where the error is.

 

[simonak]

The molar fraction is the (number of mol for the freon) / (total number of mol). I found the number of mol for the freon = 7,51161 mol.

Now I need the total number of mol. So I need the
(number of mol for the freon) + (number of mol for the azote) / (Molar weight of freon) + (Molar weight of azote).

The only thing that is missing is the (number of mol for the azote). And I found 2 ways to find it

1) volumic weight * Volume
2) PV=nRT ===> n = PV/RT ==> M = m/n ==> m = n*M ...

In each case, I don't have the volume of the azote so I cannot continue

 

[Gokul43201]

2) PV=nRT ===> n = PV/RT ==> M = m/n ==> m = n*M ...

In each case, I don't have the volume of the azote so I cannot continue

Initially, you are told that the entire volume of the tank (ie : 60 gall) contains azote (or Nitrogen) at some P, T. So, the volume of nitrogen is 60 gall. From this you can calculate n as you've proposed above.