Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
of ethanoic acid () is dissolved in of benzene. The freezing point falls by . For benzene, . Find the molar mass from the freezing point depression, the van’t Hoff factor , and the percentage of the acid that is present as dimer. The dimer forms by .
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 CO, which adds one carbon. Two hydrogen bonds join acid molecules in a dimer, so the molar mass found in benzene is near double.
Have you tried it?
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. . The true molar mass of the monomer is , so this is close to twice as much.
Step 2. The factor . If there were no pairing, . So .
Step 3. The dimer. , so . This is .
Answer: , and as dimer.
Check. If every molecule were paired, would be , and the molar mass would be .
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