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What is the molar mass of lauric acid (monoprotic) if 14.06 mL of 0.355 M KOH...

What is the molar mass of lauric acid (monoprotic) if 14.06 mL of 0.355 M KOH is needed to neutralize 1.000 g of the acid.

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Answer #1

from the given data,

    1 mol lauric acid = 1 mol KOH

no of mol of KOH reacted = M*V

                         = 0.355*14.06/1000

                        = 0.005 mol

no of mol of lauric acid = 0.005 mol

molarmass of lauric acid = w/n

                         = 1/0.005

                         = 200 g/mol

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