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?

Step-by-step explanation:

(a)   The given data is as follows.

      = 511 torr,       = 150 torr

Mass of pentane = volume × density

                            =

                            = 1265.75 g

Mass of hexane = volume × density

                          =

                          = 2279.7 g

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

No. of moles of pentane =        

                            =

                            = 17.58 mol

No. of moles of hexane =        

                            =

                            = 26.51 mol

Tota number of moles = 17.58 mol + 26.51 mol

                                     = 44.09 mol

Mole fraction of pentane () =

                 =

                 = 0.39

Mole fraction of hexane () =

                 =

                 = 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

                  =

                  = (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.

           =

           =

           =

Hence, mole fraction of pentane in the vapor that is in equilibrium with this solution is .

answer: 270

step-by-step explanation:

you have to times 200 by 1.35 to get your answer