Chemistry, 18.02.2020 18:50, mathman783

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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answered: cbbentonam72

The mass percent of Pentane in solution is 16.49% The mass percent of Hexane in solution is 83.51%

Explanation:

Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane. Convert these values to mol% using their molecular weights:

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

From Raoult's law we know:

Pp = xp $\times$ Pp - vap = yp $\times$ Pt (1)

Ph = xh $\times$ Ph - vap = yh $\times$ Pt (2)

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

(1 - xp) $\times$ Ph - vap = (1 - yp) $\times$ Pt (3)

Substituting (1) into (3) we get:

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

Rearrange for xp:

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

Subbing in the values we find:

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 $\times$ 72.15 g/mol

= 13.766 g

mh = 0.8092 mol $\times$ 86.18 g/mol

= 69.737 g

Mass% of Pentane solution = 13.766/(13.766+69.737)

= 16.49%

Mass% of Hexane solution = 83.51%

answered: Guest

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

answered: Guest

Ais the correct answer