Problem#1 : (Sawyer)Benzene, Toluene, Ethyl Benzene, and Xylene (BTEX) is the co

Question

Problem#1 : (Sawyer)Benzene, Toluene, Ethyl Benzene, and Xylene (BTEX) is the common constituent of gasoline. Vapor pressures of the pure liquids are, 0.126, 0.0380, 0.0126, and 0.0117 atm. at 25oC. Assuming an equimolar mixture of these liquids obeys Raoult's law, calculate the vapor pressure exerted by each chemical and the total vapor pressure exerted by mixture. ·If the mixture of BTEX is in equilibrium with water, estimate the solubility of benzene in water if the Henry's law constant for benzene at 25oC is 0.18 M/atm?

Explanation / Answer

We can calculate the vapour pressure of each chemical by,

benzene = P= P0benzene* n benzene, where ( p0= pressure at 25 0 c and n= moles),

P= (95.76)* (0.1 mole) (0.126 atm converted to torr),

Pbenzene= 9.576 torr (0.0126 atm),

Ptoluene =P0 Tol* n toluene,

P= 28.88* 0.1 (MOLES TAKEN AS 0.1 MOL),

P tol= 2.888 torr, (0.0038 atm)

Pethyl benzene= p0 E* n of E,

P= 9.576*0.1 =0.9576 TORR, (0.00126 atm)

Pxylene= P0 X* nX,

p= 8.892* 0.1 =0.8892 torr, (0.00117 atm)

vapour pressure of mixture can be calculated as, FOR EQUIMOLAR QUANTITY

PBTEX= (PB)*(nB)+(pT)*(TE)+(PE)*(nE)+ (PX)*(nX),

PBTEX= (95.76)*(0.1)+(28.88)*(0.1)+(9.576)*(0.1)+(8.892)*(0.1),

PBTEX= 9.576+2.888+0.9576+0.8892,

vapour pressure of mix is PBTEX=14.31 TORR, (0.01882 atm)

equimolar qty= mix=0.4 mole = water=0.4 mol,

P= K*C,

14.31= K*0.18 Mol/atm, c=0.18 mol/atm

k= 14.31 torr/0.18 mol/atm,

k=0.001882 atm/ 0.18= 0.1045 m/atm

0.1045/4=0.02613  L M/atm SOLUBLITY OF BENZENE,

from 0.1 MOLE OE BENZENE only 0.002613 mole is soluble in water

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