Respuesta :
Step-by-step explanation:
(a) The given data is as follows.
[tex]P^{o}_{1}[/tex] = 511 torr, [tex]P^{o}_{2}[/tex] = 150 torr
Mass of pentane = volume × density
= [tex]25 ml \times 50.63 g/mL[/tex]
= 1265.75 g
Mass of hexane = volume × density
= [tex]45 ml \times 50.66 g/ml[/tex]
= 2279.7 g
Now, we will calculate the moles of both pentane and hexane as follows.
No. of moles of pentane = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{1265.75 g}{72 g/mol}[/tex]
= 17.58 mol
No. of moles of hexane = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{2279.7 g}{86 g/mol}[/tex]
= 26.51 mol
Tota number of moles = 17.58 mol + 26.51 mol
= 44.09 mol
Mole fraction of pentane ([tex]X_{1}[/tex]) = [tex]\frac{moles}{\text{total moles}}[/tex]
= [tex]\frac{17.58 mol}{44.09 mol}[/tex]
= 0.39
Mole fraction of hexane ([tex]X_{2}[/tex]) = [tex]\frac{moles}{\text{total moles}}[/tex]
= [tex]\frac{26.51 mol}{44.09 mol}[/tex]
= 0.60
So, we will calculate the vapor pressure of the solution as follows.
[tex]P_{solution} = x_{1}P^{o}_{1} + x_{2}P^{o}_{2}[/tex]
where, [tex]x_{1}[/tex] = mole fraction of solution 1
[tex]P^{o}_{1}[/tex] = partial pressure of pure solvent of solution 1
[tex]x_{2}[/tex] = mole fraction of solution 2
[tex]P^{o}_{2}[/tex] = partial pressure of pure solvent of solution 2
[tex]P_{solution} = x_{1}P^{o}_{1} + x_{2}P^{o}_{2}[/tex]
= [tex]0.39 \times 511 + 0.60 \times 150[/tex]
= (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.
= [tex]\frac{\text{Partial pressure of pentane}}{\text{total pressure}}[/tex]
= [tex]\frac{0.39}{289.29}[/tex]
= [tex]1.34 \times 10^{-3}[/tex]
Hence, mole fraction of pentane in the vapor that is in equilibrium with this solution is [tex]1.34 \times 10^{-3}[/tex].
