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For the given picture:
Moles of NaOH = A × 0.04183 moles
Moles of HC2H3O2 neutralized = (A × 0.04183) moles
Mass of HC2H3O2 = (2.0598 × A)g
Percent (m/v) HC2H3O2 in vinegar = (411.96 × A)%
<h3>What is molarity?</h3>A mole is the unit of measurement used for chemicals, hence the term "molarity". Molarity, also known as solution molarity, is a technique for calculating the amount of a substance in a particular chemical solution.
Let molarity of NaOH be A from picture, volume of vinegar used is 5ml.
Now, average volume of NaOH used for titration = 0.04183 L (as per picture)
Molarity of NaOH = A
So,
(C7) Moles of NaOH = A × 0.04183
as 1 mol NaOH neutralizes 1 mol CH3COOH
(A × 0.04183) moles will neutralize (M × 0.04183) moles CH3COOH
(C8) moles of HC2H3O2 neutralized = (A × 0.04183)
Molar mass of HC2H3O2 is 60 gm/mol
So mass of (M × 0.04183) moles is (A × 0.04183 × 60) = (2.0598 × A)
(C9) Grams HC2H3O2 = (2.0598 × A)
Now volume of (vinegar+acetic acid) that was titrated = 5ml or 0.005 L
Now, M₁V₁ = M₂V₂
where,
M₁ = A
V₁ = 0.04183L
M₂ = ?
V₂ = 0.005L
put all values, value of V must be in liters
M₂ = (0.04183 × A)/0.005 = 8.366A Molar
(C10) molarity of HC2H3O2 = 8.366A Molar
Now grams of HC2H3O2 is taken from C9 and volume of vinegar is 0.005L
(C11)so mass of acetic acid/volume of vinegar(m/v)
= [(2.0598 × A)/0.005]
= (411.96 × A)%
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