3 A 25.0 cm3 portion of vinegar was transferred to a 100 cm3 volumetric flask and the solution made up to the mark with deionised water. 10.0 cm3 portions of this diluted vinegar required on average 13.60 cm3 of 0.200 mol dm−3 sodium hydroxide for neutralisation.
The equation for the reaction is:
CH3COOH NaOH → CH3COONa H2O
a Use the results to calculate the concentration of ethanoic acid in mol dm–3 of the diluted vinegar. (3 marks)
b Hence calculate the concentration of ethanoic acid in mol dm–3 of the undiluted vinegar. (2 marks)
c The density of this vinegar is 1.06 g cm–3. Use your answer to part b to calculate the concentration of ethanoic acid as a percentage by mass.
can u help me with these questions
- Thread Starter
- 25-10-2017 16:25
- Official Rep
- 27-10-2017 23:16
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