Benzoic acid (C6H5COOH, MW = 122.12 g/mol) is a non-volatile solid which dissolv
ID: 899856 • Letter: B
Question
Benzoic acid (C6H5COOH, MW = 122.12 g/mol) is a non-volatile solid which dissolves readily in benzene (C6H6), MW = 78.11 g/mol). The vapor pressure of pure benzene is 400 torr (0.533 bar) at 60 degrees C. Assume that Raoult's Law holds for solutions of benzoic acid in benzene.
A.) Consider a solution of 0.628 g of benzoic acid dissolved in 48.448 g of benzene. Calculate the vapor pressure of benzene over this solution, assuming that the only species present in the solution are C6H6 and C6H5COOH.
B.) Actually, benzoic acid dimerizes to a certain extent in benzene solution,
2C6H5COOH <--> (C6H5COOH)2
This causes the actual vapor pressure of benzene above the solution described in Part A to be 398 torr, different from the result you shoudl have obtained in Part A. On the basis of this information, calculate the number of moles of benzoic acid dimer (nD) and the number of moles of benzoic acid monomer (nM) present in this solution.
Explanation / Answer
moles of benzene = 48.448/78.11 = 0.62 moles
moles of benzoic acid = 0.628/122.12= 0.00514 moles
total number of moles = 0.6251 moles
mole fraction of benzene = 0.62/0.6251 = 0.992
Vapor pressure of the benzene = mole fraction * vapor pressure of pure benzene
= 0.992 * 400 = 396.74 torr
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Vapor pressure of the benzene = mole fraction * vapor pressure of pure benzene
398 = mole fraction *400
molefraction = 0.995
mole of benzene/mole of (benzene + benzoic acid ) = 0.995
moles of benzoic acid = 0.0049 moles
number of moles of monomer = 0.0049 moles
number of moles of dimer = 0.00514-0.0049 =2.4 *10-4
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