Step-by-step explanation:
(a) The given data is as follows.
= 511 torr,
= 150 torr
Mass of pentane = volume × density
= ![25 ml \times 50.63 g/mL](https://tex.z-dn.net/?f=25%20ml%20%5Ctimes%2050.63%20g%2FmL)
= 1265.75 g
Mass of hexane = volume × density
= ![45 ml \times 50.66 g/ml](https://tex.z-dn.net/?f=45%20ml%20%5Ctimes%2050.66%20g%2Fml)
= 2279.7 g
Now, we will calculate the moles of both pentane and hexane as follows.
No. of moles of pentane =
= ![\frac{1265.75 g}{72 g/mol}](https://tex.z-dn.net/?f=%5Cfrac%7B1265.75%20g%7D%7B72%20g%2Fmol%7D)
= 17.58 mol
No. of moles of hexane =
= ![\frac{2279.7 g}{86 g/mol}](https://tex.z-dn.net/?f=%5Cfrac%7B2279.7%20g%7D%7B86%20g%2Fmol%7D)
= 26.51 mol
Tota number of moles = 17.58 mol + 26.51 mol
= 44.09 mol
Mole fraction of pentane (
) = ![\frac{moles}{\text{total moles}}](https://tex.z-dn.net/?f=%5Cfrac%7Bmoles%7D%7B%5Ctext%7Btotal%20moles%7D%7D)
= ![\frac{17.58 mol}{44.09 mol}](https://tex.z-dn.net/?f=%5Cfrac%7B17.58%20mol%7D%7B44.09%20mol%7D)
= 0.39
Mole fraction of hexane (
) = ![\frac{moles}{\text{total moles}}](https://tex.z-dn.net/?f=%5Cfrac%7Bmoles%7D%7B%5Ctext%7Btotal%20moles%7D%7D)
= ![\frac{26.51 mol}{44.09 mol}](https://tex.z-dn.net/?f=%5Cfrac%7B26.51%20mol%7D%7B44.09%20mol%7D)
= 0.60
So, we will calculate the vapor pressure of the solution as follows.
where,
= mole fraction of solution 1
= partial pressure of pure solvent of solution 1
= mole fraction of solution 2
= partial pressure of pure solvent of solution 2
![P_{solution} = x_{1}P^{o}_{1} + x_{2}P^{o}_{2}](https://tex.z-dn.net/?f=P_%7Bsolution%7D%20%3D%20x_%7B1%7DP%5E%7Bo%7D_%7B1%7D%20%2B%20x_%7B2%7DP%5E%7Bo%7D_%7B2%7D)
= ![0.39 \times 511 + 0.60 \times 150](https://tex.z-dn.net/?f=0.39%20%5Ctimes%20511%20%2B%200.60%20%5Ctimes%20150)
= (199.29 + 90) torr
= 289.29 torr
Therefore, vapor pressure of this solution is 289.29 torr.
(b) Mole fraction of pentane in the vapor that is in equilibrium with this solution is calculated as follows.
= ![\frac{\text{Partial pressure of pentane}}{\text{total pressure}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BPartial%20pressure%20of%20pentane%7D%7D%7B%5Ctext%7Btotal%20pressure%7D%7D)
= ![\frac{0.39}{289.29}](https://tex.z-dn.net/?f=%5Cfrac%7B0.39%7D%7B289.29%7D)
= ![1.34 \times 10^{-3}](https://tex.z-dn.net/?f=1.34%20%5Ctimes%2010%5E%7B-3%7D)
Hence, mole fraction of pentane in the vapor that is in equilibrium with this solution is
.