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wel
3 years ago
11

Assuming that the density of vinegar is 1.005 g/mL, calculate the molatrity of acetic acid in vinegar from an average of 4.6% fo

r the mass percentage of acetic acid in vinegar.
Chemistry
1 answer:
posledela3 years ago
5 0
Acetic acid is 60.05 grams/mole. In 1 liter of vinegar or 1000 ml  there would be 0.046% of acetic acid = 46 ml x 1.005g/ml = 46.23 grams/60.05 grams= 0.77 moles per litre of vinegar.This then would be the concentration of acetic acid in for example 1 liter of vinegar. 
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A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o
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<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>
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2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>
  • Number of moles = Mass ÷ Molar mass
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Moles of NaOH = 13.20 g ÷ 40.0 g/mol

                          = 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
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  • Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
  • Thus, Moles of H₂SO₄ = moles of NaOH × 2

                                    = 0.33 moles × 2

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<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
  • When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
  • That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

                                = 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

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