a solution of butanoic acid and water (81.6% water by mass) is heated to 99.4 o

Question

a solution of butanoic acid and water (81.6% water by mass) is heated to 99.4oC. the vapor pressures of the pure substances at this temperature are 95 torr and 744 torr, respectively. you should assume that the acid does not dissociate appreciably under these conditions.

(a.) determine the composition (mole fractions) of this solution.

(b.) assuming the vapor behaves ideally, calculate the composition of the vapor in equilibrium with this solution.

(c.) this solution is actually an azeotrope that boils at 99.4oC. Calculate the activity coefficients for the two species.

Explanation / Answer

Ans:

(a) Mass of water = 81.6 g and its molar mass = 18 g/mol

No.of moles of water, nwater = 81.6/18 = 4.53 mol

Mass of butanoic acid = 100 - 81.6 = 18.4 g and its molar mass = 88 g/mol

No. of moles of butanoic acid, nbutanoic acid = 18.4/88 = 0.21 mol

Therefore mole fraction of water = nwater/(nwater + nbutanoic acid)

Xwater = 4.53 / (4.53 + 0.21)

Xwater = 0.96

Hence, Xbutanoic acid = 1 - 0.96 = 0.04

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