The vapor pressure of pure benzene (in bar) is given by the Antoine equation, th

Question

The vapor pressure of pure benzene (in bar) is given by the Antoine equation, that has the form log_10(P) = A -[B/(T + C)] with the coefficients, A = 4.72583, B = 1660.7 and C = 1.461 and T is the absolute temperature in Kelvins Calculate the vapor pressure of the pure solvent at 330 K. Enter your answer to the nearest Pascal; e.g., 19322 Pa. A solution of this compound at 330 K, containing a small amount of nonvolatile solute has a vapor pressure of 46000 Pa. What is the boiling point elevation (Delta T_b) of this solution, given that the ebullioscopic constant is 2.53 degree C kg/mol. Enter your answer to the nearest one hundredth of a degree; e.g. 0.17 degree C.

Explanation / Answer

At 330K, logP:= 4.72583- 1660.7/(330-1.461) =-0.33

P = 0.467 bar =0.467*105pa = 46700 pa

Vapor pressure of solution = mole fraction of Benzene* Pure component vapor pressure

46000 = 46700* mole fraction benzene

Mole fraction benzene = 46000/46700=0.985

Basis : 1mole of mixture. This contains 0.985 moles Benzene and 1-0.985= 0.015 moles of nonvolatile solute. Mass of Benzene = moles* Molar mass of Benzene = 0.985*78 = 76.83gm = 76.83/1000 kg

Molality = moles of solute/ kg of benzne = 0.015*1000/76.83=0.195 mole/Kg

Boiling point elevation = i*kb*m = 1*2.53*0.195 =0.495 deg.c

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