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Topic: Mole fraction of chloroform  (Read 5075 times)

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Offline Gobo

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Mole fraction of chloroform
« on: January 30, 2012, 09:10:03 PM »
A solution is prepared from equal masses of these two compounds at this temperature.
Calculate the mole fraction of the chloroform in the vapor above the solution.
Given Information:
P[CCl4] = 0.526 atm
P[CHCl3] = 0.354 atm

^ both of these compounds are volatile, so they both contribute to the VP of the sol'n.

Attempt:
Let x rep. # of grams of CCl4.

mol CCl4 = (x g CCl4)*(1 mol CCl4)/(153.81g CCl4) = (1/153.81)*x mol CCl4
mol CHCl3 = (x g CHCl3)*(1 mol CHCl3)/(119.368g CHCl3) = (1/119.368)*x mol CHCl3

Now, this is what I did:
I found the vapour pressures of each of them individually using Rauolt's Law. Then I added those vapour pressures up together to get the total vapour presure.

I found the mole fraction of chloroform to be 0.563 while I was getting the vapour pressure above solution for chloroform alone. This was prior to me adding them up to get the total pressure of the solution. Then, I set up the equation somewhat like this:

P[total] = P[CHCl3] + P[CCl4]
P[total] = P_[CHCl3]*(X_CHCl3) + P[CCl4]
P[total] - P[CCl4]
----------------
P_[CHCl3]

= X_CHCl3

^ So what I mean by P_[CHCl3] is the pressure of chloroform that was given to us in the question. Now, I solve for mole fraction like this and I get the same exact mole fraction I did from before,

0.563.

However, my electronic assignment is saying that this is incorrect. Why?

Offline Borek

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Re: Mole fraction of chloroform
« Reply #1 on: January 31, 2012, 05:40:08 AM »
0.563 is a correct value of molar fraction of chloroform in the solution.

What is a partial pressure of the CHCl3 above the solution?

What is a partial pressure of the CCl4 above the solution?
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