If 2.50 ml of vinegar requires 34.50 ml of 0.0960 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in 500.0 ml of this vinegar? Density of vinegar=1.01 g ml

The mass of vinegar present in 500mL of solution is 39.6 g.

Concentration of acid (CA) = ?

Volume of acid(VA) = 2.50 ml

Concentration of base (CB) =0.0960 M

Volume of base (VB) =  34.50 ml

Number of moles of acid = 1

Number of moles of base = 1

Equation of the reaction;

CH3COOH (aq) + NaOH(aq) > CH3COONa(aq) + H2O(l)

From CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA = CBVBNA/VANB

CA = 0.0960 × 34.50 × 1/2.50 × 1

CA = 1.32 M

Number of moles = concentration  × volume

Number of moles = 1.32 M × 500/1000

Number of moles = 0.66 moles

Number of moles = mass/molar mass

Mass = Number of moles  × molar mass

Molar mass of acetic acid = 60 g/mol

Mass =  0.66 moles  ×  60 g/mol

Mass = 39.6 g

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Grams of acetic acid in 500 ml vinegar=39.744 g

Titration (equivalence point⇒mol acid(CH₃COOH)=mol base(NaOH))

M₁V₁n₁=M₂V₂n₂(n=valence acid/base, for CH₃COOH and NaOH=1)

For MW CH₃COOH = 60 g/mol, mass of acetic acid in 1 L :

for 500 ml :

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

16

explanation:

the marks go by 4s