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.

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bhuvna789456 bhuvna789456 · Answer 1 · 2020-02-25T07:47:32+01:00

Explanation:

Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.

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 [tex]\times[/tex] Pp - vap = yp [tex]\times[/tex] Pt (1)

Ph = xh [tex]\times[/tex] Ph - vap = yh [tex]\times[/tex] Pt (2)

Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:

(1 - xp) [tex]\times[/tex] Ph - vap = (1 - yp) [tex]\times[/tex] Pt (3)

(1-xp) [tex]\times[/tex] Ph - vap = (1 - yp) [tex]\times[/tex] xp [tex]\times[/tex] 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%

Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:

mp = 0.1908 mol [tex]\times[/tex] 72.15 g/mol

= 13.766 g

mh = 0.8092 mol [tex]\times[/tex] 86.18 g/mol

= 69.737 g

= 16.49%