Respuesta :
First we need to calculate the number of moles of NaOH titrated.
molar concentration = number of moles / solution volume (liter)
number of moles = molar concentration × solution volume
number of moles of NaOH = 1.5 × 0.0075 = 0.01125 moles
Then we look at the chemical reaction:
CH[tex]_{3}[/tex]-COOH + NaOH = CH[tex]_{3}[/tex]COONa + H[tex]_{2}[/tex]O
We can see that 1 mole of acetic acid is reacting with one mole of sodium hydroxide. Then we can conclude that 0.01125 moles of sodium hydroxide reacts with 0.01125 moles of acetic acid.
Now we can get the mass of acetic acid:
number of moles = mass (grams) / molecular mass (g/mol)
mass = number of moles × molecular mass
mass of acetic acid = 0,01125 × 60 = 0.675 g
We assume that the density of the vinegar = 1 g/mL, so the mass percent of acetic acid is:
concentration of acetic acid = (mass of acetic acid / mass of vinegar) × 100
concentration of acetic acid = (0.674 / 7) × 100 = 9.6 %
The mass percent of acetic acid in the vinegar sample 9.6 %.
The reaction will be,
[tex]\rm \bold { CH_3-COOH + NaOH \leftrightharpoons CH_3COONa + H_2O}[/tex]
Means 1 mole acetic acid is reacting with one mole of sodium hydroxide.
Number of moles of NaOH titrated,
Number of moles of [tex]\rm \bold{ NaOH = 1.5 \times 0.0075 = 0.01125 moles}[/tex]
number of moles of [tex]\rm \bold { CH_3COOH}[/tex] = 0.01125 moles
Mass of [tex]\rm \bold { CH_3COOH}[/tex] = [tex]\rm \bold{ 0.01125 \times 60 = 0.675 g}[/tex]
Assume that the density of the vinegar is 1 g/mL
The mass percent of acetic acid can be calculated,
The mass percent of acetic acid = (mass of acetic acid / mass of vinegar) 100
The mass percent of acetic acid = [tex]\rm \bold{\frac{0.674}{7} \times 100 = 9.6 } \\[/tex]
Hence, we can conclude that the mass percent of acetic acid in the vinegar sample 9.6 %.
To know more about mass percent, refer to the link:
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