Question

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?

Answer 1

Step-by-step explanation:

(a)   The given data is as follows.

     $P^{o}_{1}$ = 511 torr,      $P^{o}_{2}$ = 150 torr

Mass of pentane = volume × density

                            = $25 ml \times 50.63 g/mL$

                            = 1265.75 g

Mass of hexane = volume × density

                          = $45 ml \times 50.66 g/ml$

                          = 2279.7 g

Now, we will calculate the moles of both pentane and hexane as follows.

No. of moles of pentane = $\frac{mass}{\text{molar mass}}$        

                            = $\frac{1265.75 g}{72 g/mol}$

                            = 17.58 mol

No. of moles of hexane = $\frac{mass}{\text{molar mass}}$        

                            = $\frac{2279.7 g}{86 g/mol}$

                            = 26.51 mol

Tota number of moles = 17.58 mol + 26.51 mol

                                     = 44.09 mol

Mole fraction of pentane ($X_{1}$) = $\frac{moles}{\text{total moles}}$

                 = $\frac{17.58 mol}{44.09 mol}$

                 = 0.39

Mole fraction of hexane ($X_{2}$) = $\frac{moles}{\text{total moles}}$

                 = $\frac{26.51 mol}{44.09 mol}$

                 = 0.60

So, we will calculate the vapor pressure of the solution as follows.

$P_{solution} = x_{1}P^{o}_{1} + x_{2}P^{o}_{2}$

where,   $x_{1}$ = mole fraction of solution 1

              $P^{o}_{1}$ = partial pressure of pure solvent of solution 1

             $x_{2}$ = mole fraction of solution 2

             $P^{o}_{2}$ = partial pressure of pure solvent of solution 2

      $P_{solution} = x_{1}P^{o}_{1} + x_{2}P^{o}_{2}$

                  = $0.39 \times 511 + 0.60 \times 150$

                  = (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.

           = $\frac{\text{Partial pressure of pentane}}{\text{total pressure}}$

           = $\frac{0.39}{289.29}$

           = $1.34 \times 10^{-3}$

Hence, mole fraction of pentane in the vapor that is in equilibrium with this solution is $1.34 \times 10^{-3}$.