Question

If the vapor pressure of pure benzene is 96.1 mm Hg, what must the vapor pressure of pure toluene be in order for the 50/50 % mixture of benzene and toluene have a total vapor pressure of 63.2 mm Hg?

Answer 1

`P_(tol)=30.34mmHg`

Step-by-step explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the vapor pressure of pure toluene by using the Raoult's law as shown below:

`y_(tol)P_(mix)=x_(tol)P_(tol)\\\\y_(ben)P_(mix)=x_(ben)P_(ben)`

Thus, we solve for the mole fraction of benzene in the vapor phase first:

`y_(ben)=(x_(ben)P_(ben))/(P_(mix)) =(0.5*96.1mmHg)/(63.2mmHg)=0.76`

Which means that the mole fraction of toluene in the vapor phase is 0.24, and therefore, the vapor pressure of pure toluene turns out to be:

`P_(tol)=(y_(tol)P_(mix))/(x_(tol)) =(0.24*63.2mmHg)/(0.5)=30.34mmHg`

Regards!