The vapor above a mixture of pentane and hexane at room temperature contains 35.5% pentane by mass. What is the mass percent composition of the solution? Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature.

Explanation:

Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol

Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol

Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%

Hexane mol%: yh = 100 - 39.68 = 60.32%

Pp-vap = 425 torr = 0.555atm

Ph-vap = 151 torr = 0.199atm

Pp = xp Pp - vap = yp Pt                                       (1)

Ph = xh Ph - vap = yh Pt                                       (2)

(1 - xp) Ph - vap = (1 - yp) Pt                                   (3)

(1-xp) Ph - vap = (1 - yp) xp Pp - vap / yp            (4)

xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap)      (5)

Pentane mol% in solution: xp = 19.08%  

Hexane mol% in solution: xh = 80.92%

mp = 0.1908 mol 72.15 g/mol

     = 13.766 g

mh = 0.8092 mol 86.18 g/mol

     = 69.737 g

                                                       = 16.49%

i think it is d. e because the next one is igneous.

1 is the answer to this problem