Chem 1120 Homework 3 Name 13. Pentane (C,Hsa) and hexane (CaHja) form an ideal s

Question

Chem 1120 Homework 3 Name 13. Pentane (C,Hsa) and hexane (CaHja) form an ideal solution. At 25"C, the vapor pressures of pentane and hexanes are 511 and 150. torr, respectively. A solution is prepared by mixing 25 mL pentane (density 0.63 g/mL) with 45 mL hexane (density = 0.66 g/mL). a. What is the vapor pressure of the resulting solution? b. What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution. 14. what is the composition of a methanol (CH,0propanol (CH,CHCHO) solution that have a vapor pressure of 174 torr at 40°C? At 40"C, the vapor pressures of pure methanol and pure propanol are 303 and 44.6 torr, respectively. Assume the solution is ideal.

Explanation / Answer

Molar mass of pentane(C5H12)= 72 g/mole and molar mass of hexane= 86 g/mole

Mass of pentane= density* volume =0.63*25 gm=15.75 gm

Mass of hexane= density* volume = 0.66*45 gm=29.7 gm

Moles= mass/moalr mass

Moles : Pentane =15.75/72 =0.22, Hexane= 29.7/86=.0.345

Total moles = moles of pentane+ moles of hexane=0.22+0.345=0.565

Let x1= mole fraction of pentane, and x2= mole fraction of hexane

Mole fraction =moles of component/total moles

Hence x1= 0.22/0.565=0.39 and x2=1-0.39=0.61

Let P1sat= pure component vapor pressure of pentane= 511 torr and

P2sat= pure component vapor pressure of hexanee =150 torr

From Raoult’s law, y1P= x1P1sat and y2P= x2P2sat where y1,y2 are mole fractions of pentane and hexane in the vapor phase. P is the total pressure, the vapor pressure of resulting solution.

Since x1+x2=1 and y1+y2=1

Hence y1P+y2P= x1P1sat+x2P2sat, or P =x1P1sat+x2P2sat=0.39*511+0.61*150 Torr=290.79 Torr

Now y1 =x1P1sat/P= 0.39*511/290.79=0.68

2.

From Raoult’s law, y1P= x1P1sat and y2P= x2P2sat where y1,y2 are mole fractions of methanol and propanol in the vapor phase. P is the total pressure, the vapor pressure of resulting solution.x1,x2 are mole fractions of methanol and propanol in the liquid phase. P1sat and P2sat are pure component vapor pressures of methanol and propanol.

x1+x2= 1 and y1+y2=1

P1sat= 303 torr and P2sat= 44.6 Torr, P= 174 torr

Hence y1P+y2P= x1P1sat+x2P2sat, P= x1P1sat+(1-x1)P2sat= 174

Hence x1*(P1sat-P2sat)= 174-44.6=129.4

Hence x1= 129.4/(303-44.6)= 0.5

Hence from y1P=x1P1sat, y1= x1P1sat/P= 0.5*303/174= 0.87, x2=1-x1=1-0.5=0.5 and y2= 1-0.87=0.13

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