Q46 · Practice questionComplex familiar5 marks
QUESTION 46 (5 marks)
A vinegar sample is diluted by transferring 10.00 mL to a 100.0 mL volumetric flask. A 20.00 mL aliquot of the dilution requires 16.40 mL of 0.1000 mol L⁻¹ NaOH. Treat all acidity as ethanoic acid; its molar mass is 60.05 g mol⁻¹.
a) Calculate the original acid concentration in mol L⁻¹. [3 marks]
b) Convert the result to grams of ethanoic acid per 100 mL of vinegar. [2 marks]
WORKED SOLUTION
5 marksPractice marking scheme
a) 0.8200 mol L⁻¹. b) 4.924 g per 100 mL.
. Diluted concentration is 0.08200 mol L⁻¹; original is ten times greater. Mass in 0.100 L is g.
- Calculates titrant/acid moles. [1 marks]
- Calculates diluted concentration. [1 marks]
- Applies the tenfold dilution factor. [1 marks]
- Uses mass = cVM. [1 marks]
- Calculates 4.924 g per 100 mL. [1 marks]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabusHow many marks did you earn?
Compare your working with the guide above.