When 0.72 g of acetic acid is dissolved in 25.0 gof benzene, the freezing point

Question

When 0.72 g of acetic acid is dissolved in 25.0 gof benzene, the freezing point of the solution is4.25°C. a. Calculate the apparent molecular weight ofacetic acid in benzene.
b. Your answer to part “a” should be differentfrom normal molecular weight of acetic                acid. Explain the cause of this difference. a. Calculate the apparent molecular weight ofacetic acid in benzene.
b. Your answer to part “a” should be differentfrom normal molecular weight of acetic                acid. Explain the cause of this difference.

Explanation / Answer

molality = 0.72 g acetic acid /MW /(0.025 kg benzene) = 28.8 /MW delT = Kf*m delT = 5.5 - 4.25 = 1.25 C (5.5 is the normal freezing pointof benzene) Kf = 4.90 1.25 = 4.90 *28.8/MW MW = 112.89 g/mol b) acetic acid ionizies. delT = Kf*m, where m is the molalityof ions/total solutes. So acetic acid exists as CH3COO- and H+ ionsand some CH3COOH. Hence m ~ twice as much as 28.8/MW. The answer is good, because MW of acetic acid is actually about 60g/mol. Here we calculate almost twice that number. The reason isthat the molality of ions/total solutes is just slightly higher.This is the effect of the van't Hoff factor. delT = Kf*m *i , where i = van't hoff vactor, 1

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