Question

Moles of NaOH used in titration mole NaOH Moles of HC2H3O2 neutralized by NaOH mole HC2H3O2 Molarity of HC2H3O2 M HC2H3O2 Grams HC2H3O2 g HC2H3O2 Percent (m/v) HC2H3O2 in vinegar % HC2H3O2

Answer 1

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)%

What is molarity?

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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