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Percentages of Solutions

Submitted by mrkegley on Thu, 08/06/2009 - 11:45

I have tried many different approaches to this problem and none of the answers match my choices, so I'm obviously doing something wrong please help!  Telling me what equation to use and how to solve will help me out greatly!

A solution of Carbon Tetrachloride in benzene, C6H6, at 20 degrees Celsius has a total vapor pressure of 78.5 mmHg.  (assume that this solution is ideal)  The vapor pressure of pure benzene at this temperature is 74.61 mmHg and its density is 0.87865 g/cm3; the vapor of pure carbon tetracholride is 91.32 mmHg and it's density is 1.5940 g/cm3.  What percentage of the volume of this solution is due to carbon tetrachloride? (Hint: assume that you have 1.000 L of solution.)

A) 75.2%
B) 76.7%
C) 14.3%
D) 24.8%
E) 23.3%

Thank you!

Is the answer 24.82%
I am not an expert but if it is, let me know and I'll guide you.

Just in case the answer is right, here is my method

I used the equation
VPsolution = VP1 x mole fraction1 + VP2 x mole fraction2
where VP1 = VP of pure component1 etc
I let the mole fraction of benzene = y, so that m.f. CCl4 = 1-y
Plugging in I got
78.5 = 74.61y + 91.32(1-y)
78.5 = 74.61y + 91.32 - 91.32y
This gives y = 0.7672 as mole fraction of benzene.
Mole fraction of CCl4 = 0.2328
As these are mole ratios, I found the mass of these mole values using the mass of 1 mole of each. I use simple values for relative atomic mass (you use the values you are expected to use)
(C=12, H=1,Cl=35.5)
C6H6 = 78 g/mole and CCl4 = 154 g/mol
Mass of C6H6 based on mole fraction = 0.7672 x 78 = 59.8416g
Mass of CCl4 based on mole fraction = 0.2328 x 154 = 35.8512g

I then converted these masses in volumes using density (volume = mass/density)
Volume C6H6 = 59.8416/0.8786 = 68.1102 mL
Volume CCl4 = 35.8512/1.5940 = 22.4913 mL

% CCl4 = (22.4913/(68.1102 + 22.4913) x 100
            = (22.4913/90.6032) x 100
            = 24.82