This means that you used moles of naoh to titrate moles of

Reference no: EM13385204

You are asked to titrate a 6.00 ml sample of vinegar using 0.25M NaOH. The mass of your vinegar sample was 6.168g, so the density was g/ml. It took 80.18ml of your NaOH to reach the endpoint of the titration. This means that you used moles of NaOH to titrate moles of acetic acid in the vinegar sample or grams of acetic acid. The percent mass of your sample of vinegar was %.

Reference no: EM13385204

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