A solution of liquid toluene dissolved in liquid benzene has a benzene mole fraction of .850. Calculate the vapor pressure of the solution given that the vapor pressures of pure benzene and toulene are 183 mmHg and 59.2 mmHg, respectively?

I tried solving it like this: 183(.850) + 59.2(.850) = 206 mmHg but this is wrong, please explain what I am doing wrong

Mar 26, 2017

Now the sum of the mole fractions must equal $1$. Do you agree?

Explanation:

$\text{Mole fraction of benzene}$ $=$ n_"benzene"/(n_"benzene"+n_"toluene")=0.850

And given a binary solution:

$\text{Mole fraction of toluene}$ $=$ n_"toluene"/(n_"benzene"+n_"toluene")=0.150

Why? Because ${\chi}_{\text{benzene"+chi_"toluene}} = 1$

And thus $\text{vapour pressure of the solution}$ is equal to the weighted average of the mole fractions:

"Vapour pressure"=(chi_"benzene"xx183+chi_"toluene"xx59.2)*mm*Hg

$= \left(0.850 \times 183 + 0.150 \times 59.2\right) \cdot m m \cdot H g = 164.5 \cdot m m \cdot H g$

I think you will kick yourself when you realize the mistake you made. Remember, we have ALL made this sort of error. It is easy to do.

Is this answer consistent with your model answer? As usual, the vapour pressure is enriched with respect to the proportion of the more volatile component of the solution, i.e. enriched with respect to benzene.