a) Calculate the grams of acetic acid (CH 3 COOH) that would be present in one l
ID: 844007 • Letter: A
Question
a) Calculate the grams of acetic acid (CH3COOH) that would be present in one liter of your unknown solution. Use the mean value of the molarity from your sample.
b) Calculate the mass in grams of 1.000 liter of the acetic acid solution. The density of vinegar is 1.002 g/mL.
c) Calculate the mass percent CH3COOH in your sample
Trial 1
Trial 2
Trial 3
Trial 4
Volume of unknown acid solution used (mL)
10.0 mL
10.0 mL
10.0 mL
10.0 mL
NaOH buret, final volume reading (mL)
15.75 mL
8.79 mL
12.17 mL
10.06 mL
NaOH buret, initial reading (mL)
50.00 mL
49.00 mL
50.00 mL
50.00 mL
Volume of NaOH solution used (mL)
34.25 mL
40.21 mL
37.83 mL
39.94 mL
Molarity of unknown (M)
0.34 M
0.40 M
0.38 M
0.40 M
Average Molarity of vinegar solution (M)
0.38 M
Trial 1
Trial 2
Trial 3
Trial 4
Volume of unknown acid solution used (mL)
10.0 mL
10.0 mL
10.0 mL
10.0 mL
NaOH buret, final volume reading (mL)
15.75 mL
8.79 mL
12.17 mL
10.06 mL
NaOH buret, initial reading (mL)
50.00 mL
49.00 mL
50.00 mL
50.00 mL
Volume of NaOH solution used (mL)
34.25 mL
40.21 mL
37.83 mL
39.94 mL
Molarity of unknown (M)
0.34 M
0.40 M
0.38 M
0.40 M
Average Molarity of vinegar solution (M)
0.38 M
Explanation / Answer
1 L 0.38 M acetic acid contains 0.38 x 1 moles = 0.38 moles
MW of acetic acid = 60.05
So, grams of acetic acid = 0.38 x 60.05 g = 22.82 g
Mass of acetic acid = 1000 mL x 1.002 g/mL = 1002 g
So, 1002 g contains 22.82 g
So, mass percentage = 22.82 x 100 / 1002 = 2.28 %
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