Question

At 20 °C, the vapor pressure of benzene(C6H8) is 75 torr, and that of toluene(C7H8) is 22 torr. Assuming that benzene and toluene from and Idea solution.

A) What is the composition in mole fractions of a solution that has a vapor pressure of 33 torr at 20 °C
B) What is the mole fraction of benzene in the vapor above the solution described in part A?

Answer 1

Step-by-step explanation:

(a) It is known that relation between vapor pressure and mole fraction is as follows.

vapor pressure =
`x1 * P_(1) + x_(2) * P_(2)`

where,
`x_(1)` = mole fraction of component one

`P_(1)` = vapor pressure of component one when pure

Also,
`x_(1) + x_(2)` = 1

Now, putting the given values into the above formula as follows.

vapor pressure =
`x1 * P_(1) + x_(2) * P_(2)`

33 =
`x_(1) * 75 + (1 - x_(1)) * 22`

11 =
`(75 - 22) * x_(1)`

`x_(1)` = 0.20754717

= 0.21 (approx)

And,
`x_(2) = 1 - x_(1)`

= (1 - 0.21)

= 0.79

Hence, the composition in mole fractions of a solution that has a vapor pressure of 33 torr at
`20^(o)C` is 0.79.

(b) According to Raoult's law,

`x_(1) * P_(1) = (P_(vapor)) * y_(1 )`

where,
`y_(1)` = composition in gas phase

Therefore, calculate the value of
`y_(1)` (mole fraction of benzene) as follows.

`y_(1) = {x_(1) * (P_(1))/(P_(vap))`

=
`(0.21 * 75)/(33)`

= 0.471698114

= 0.47 (approx)

Thus, mole fraction of benzene in the given vapor is 0.47.