Consider Example 12.10. Suppose the experiment is repeated with 0.032 mol of helium instead with everything else staying the sam
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Answer:
5.01%
Explanation:
Density of vinegar = mass/volume
Mass of 10.00 mL = density x volume
= 1.006 x 10 = 10.06 g
From the equation of reaction:
1 mole pf CH3COOH requires 1 mole of NaOH for neutralization.
mole of NaOH = molarity x volume
= 0.5062 x 0.01658
= 0.008392796 mole
0.008392796 mole of NaOH will therefore require 0.008392796 mole of CH3COOH.
mass of CH3COOH = mole x molar mass
= 0.008392796 x 60.052
= 0.504 g
Percentage by mass of acetic acid in the vinegar = 0.504/10.06 x 100%
= 5.01%
The percent by mass of acetic acid in the vinegar is 5.01%