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  1. Problems
  2. /1.20 g of ethanoic acid (60.05 g/mol) is dissolved in 50.0 g of benzene…
HardChemistryHigh School (11–12)Organic Compounds with Functional Groups › Aldehydes, Ketones and Carboxylic AcidsNumerical

1.20 g of ethanoic acid (60.05 g/mol) is dissolved in 50.0 g of benzene…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

1.20 g1.20\ \text{g}1.20 g of ethanoic acid (60.05 g/mol60.05\ \text{g/mol}60.05 g/mol) is dissolved in 50.0 g50.0\ \text{g}50.0 g of benzene. The freezing point falls by 1.05 K1.05\ \text{K}1.05 K. For benzene, Kf=5.12 K kg/molK_f = 5.12\ \text{K kg/mol}Kf​=5.12 K kg/mol. Find the molar mass from the freezing point depression, the van’t Hoff factor iii, and the percentage of the acid that is present as dimer. The dimer forms by 2 CH3COOH⇌(CH3COOH)22\,\text{CH}_3\text{COOH} \rightleftharpoons (\text{CH}_3\text{COOH})_22CH3​COOH⇌(CH3​COOH)2​.

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g/mol
Need a reminder? The key ideaShowHide

The carboxyl carbon is carbon 1. Acids come from oxidation of primary alcohols, aldehydes and alkylbenzenes (a benzylic hydrogen is needed), from hydrolysis of nitriles, amides and esters, and from RMgX with CO2_22​, which adds one carbon. Two hydrogen bonds join acid molecules in a dimer, so the molar mass found in benzene is near double.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Molar mass. M=KfwΔT×W=5.12×1.201.05×0.0500=117 g/molM = \dfrac{K_f w}{\Delta T \times W} = \dfrac{5.12 \times 1.20}{1.05 \times 0.0500} = 117\ \text{g/mol}M=ΔT×WKf​w​=1.05×0.05005.12×1.20​=117 g/mol. The true molar mass of the monomer is 60.05 g/mol60.05\ \text{g/mol}60.05 g/mol, so this is close to twice as much.

Step 2. The factor iii. If there were no pairing, ΔT=5.12×1.20/60.050.0500=2.05 K\Delta T = 5.12 \times \dfrac{1.20/60.05}{0.0500} = 2.05\ \text{K}ΔT=5.12×0.05001.20/60.05​=2.05 K. So i=1.052.05=0.513i = \dfrac{1.05}{2.05} = 0.513i=2.051.05​=0.513.

Step 3. The dimer. 0.513=1−α/20.513 = 1 - \alpha/20.513=1−α/2, so α=2(1−0.513)=0.974\alpha = 2(1 - 0.513) = 0.974α=2(1−0.513)=0.974. This is 97.4 %97.4\ \%97.4 %.

Answer: 117 g/mol117\ \text{g/mol}117 g/mol, i=0.513i = 0.513i=0.513 and 97.4 %97.4\ \%97.4 % as dimer.

Check. If every molecule were paired, iii would be 0.50.50.5, and the molar mass would be 120 g/mol120\ \text{g/mol}120 g/mol.

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