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Pad 1:33 PM 53% chegg.com Post a question Answers from our experts fo homework questions Chapter T15, Problem IP Show all steps : Bookmark ON Enter question In the accompanying chart are approximate vapor pressures for benzene and toluene at various temperatures CONTINUE TO 20 questions ren Benzene 30 20 Toluene 30 180 270 37 140 70 550 760 1010 1340 70 My Textbook Soluti 100 100 110 CALCULUS a. What is the mole fraction of each component if 3.9 g of benzene C6Hs is dissolved in 4.6 g of toluene C7H8? b. Assuming that this mixture is ideal-that is, it follows Raoult's Law-what is the partial vapor pressure of benzene in this mixture at 50°C? c. Estimate to the nearest degree the temperature at which the vapor pressure of the solution equals 1 atm (bp of the solution) d. Calculate the composition of the vapor (mole fraction of each component) that is in equilibrium in the solution at the boiling point of this solution e. Calculate the composition in weight percentage of the vapor that is in equilibrium with the solution by Chegg by Chegg A Microscale Calc Approach...Early 5th Edition View all solutions Chegg tutors wh right now David Columbia L Step-by-step solution Rebecca University oExplanation / Answer
Molar masses ( g/mole); Benzene (C6H6) = 78 g/mole, Toluene (C7H8)= 7*12+8= 92 g/mole
Moles =mass/ molar mass
Moles : Benzene = 3.9/78=0.05, toluene = 4.6/92 =0.05
Total moles of mixture= 0.05+0.05=0.1
Mole fraction : moles of component/ total moles
Mole fraction : Benzene = toluene =0.05/0.1= 0.5
At 50 deg.c, vapor pressure (Psat); Benzene :P1sat= 270 mm Hg and toluene= 95 mm Hg
From Raoult’s law
Y1P= x1P1sat, y2P= x2p2sat, where y1, y2 are mole fraction of Benzene and toluene in the vapor phase, x1, x2 are mole fractions of benzene and toluene in the liquid phase respectively,
Partial pressure of Benzene y1P= 0.5*270= 135 mm Hg
Addition of two equation gives P= x1P1sat+x2P2sat
760= 0.5*(P1say+P2sat)
P1sat+P2sat= 760/0.5= 1520 mm Hg
So the mixture boils at a temperature at which the sum of vapor pressure equals 1520 mm Hg.
The correct temperature at which mixture boils is obtained by drawing plots of ln P Vs 1/T( T in K)
The plot of lnP vs 1/T is straight lines. From the equation of best fit, we can correctly predict the temperature at which P1sat+P2sat= 1520 mm Hg.
The plots are shown below .
from the plots, the temperature 365.125 correspond to P1sat+P2sat= 1520 mm Hg
The calculations are done in the excel and are shown
. Pressure
given P= 760 mm Hg, P1sat= 1082.27, x1=0.5, P2sat= 438.16, x2=0.5
y1= x1P1sat/P= 0.5*1082.27/760 =0.71, y2= 1-0.71=0.29
these are mole fractions which are to be converted to mass fractions.
Basis : 1 mole of vapor, moles of Benzene = 0.71, mass of Benzene= moles* molar mass=0.71*78=55.38 gm, mass of toluene= 0.29*92= 8.12, total mass of vapor =55.38+8.12 =63.5 gm
mass fractions : Benzene = 55.38/63.5 =0.87 and toluene = 1-0.87=0.13
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