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What is the average Mw for a fatty acid mixture whose fats (1 gram) required 0.0034...

What is the average Mw for a fatty acid mixture whose fats (1 gram) required 0.0034 equivalents of NaOH to titrate. Remember: 1 Fatty acid has 1 equivalent. 1. Ew = Mw (#H+) 2. Ew = mass/eq a.) 294.12 g/mol b.) 145.04 g/mol c.) 345.56 g/mol d.) 132.04 g/mol

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Answer #1

1 fatty acid has 1 equivalent carboxy group. Therefore, to titrate 1 equivalent fatty acid, 1 equivalent NaOH will be required(as acid base reactions occur in equivalence). According to the question, 0.0034 equivalent NaOH is required to titrate 1 g of fatty acid. So, based on the logic discussed above, 1 g of fatty acid must contain 0.0034 equivalents. So weight of 1 equivalent fatty acid, i.e. its equivalence weight is (1/0.0034)g = 294.12 g. Also for fatty acids, molecular weight = equivalence weight.

So, the average molecular weight is a) 294.12 g/mol

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