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neonofarm [45]
2 years ago
9

If a 7.0 mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5 mL of 1.5M NaOH, what is the mass pe

rcent of CH3COOH in the vinegar sample? Show your work.
Chemistry
1 answer:
NNADVOKAT [17]2 years ago
3 0

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_{3}-COOH +  NaOH = CH_{3}COONa + H_{2}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 %

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Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
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A=\epsilon \times C\times l

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