Answer:
<em>(a) </em>3.96 (w/w) %
<em>(b) </em>0.663M
Explanation:
The acetic acid, CH₃COOH, reacts with NaOH as follows:
CH₃COOH + NaOH → CH₃COONa + H₂O
<em>Where 1 mole of acetic acid reacts with 1 mole of NaOH</em>
<em />
In the equivalence point of the titration, the moles of acetic acid are equal to moles of NaOH added. To solve this question we need to find the moles of NaOH and thus, the mass of acetic acid
<em>Moles NaOH = Moles acetic acid:</em>
33.0cm³ = 0.033L * (0.100moles / L) = 0.0033 moles
<em>Mass acetic acid -Molar mass: 60g/mol-:</em>
0.0033 moles * (60g / mol) = 0.198g
<em>a) </em>Weight percent is 100 times the ratio of the mass of acetic acid (0.198g) in the mass of vinegar (5.00g):
0.198g / 5.00g * 100 = 3.96 (w/w) %
<em>(b) </em>The volume of the vinegar in the 5.00g is:
5.00g * (1mL / 1.005g) * (1L / 1000mL) = 0.00498L
And the molarity is:
0.0033moles / 0.00498L = 0.663M
