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)
[tex]\tt M_1\times 2.5\times 1=0.0960\times 34.5\times 1\\\\M_1(acetic~acid)=1.3248~mol/L[/tex]
For MW CH₃COOH = 60 g/mol, mass of acetic acid in 1 L :
[tex]\tt 1.3428~mol/L\times 60~g/mol=79.488~g/L[/tex]
for 500 ml :
[tex]\tt 79.488\times 0.5~L=39.744~g[/tex]
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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