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Pentane (C5H12) and hexane (C6H14) combine to form an ideal solution. At 258C the vapor pressures of pen- tane and hexane are 511 and 150. torr, respectively. A solution is prepared by mixing 25 mL of pentane (density 5 0.63 g/mL) with 45 mL of hexane (density 5 0.66 g/mL).
(a) What is the vapor pressure of this solution? _____ torr
(b) What is the mole fraction of pentane in the vapor that is in equilibrium with this solution?

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].

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