Avinegar sample was analyzed and found to be 5.88% acetic acid by volume. the density of pure acetic acid is 1.049 g/ml. what volume of 0.200m naoh would be required to titrate a 2 ml aliquot of this vinegar?

Overall equation: 
CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l) 

First determine the volume of acetic acid. If the volume of vinegar is 2 mL and 5.88% of it is acetic acid then the volume of acetic acid is 0.1176 mL. 

(5.88/100) X 2 mL = 0.1176 mL 

Now, using the d=m/v formula, we look for the mass of the acetic acid. 

d= 1.049 g/mL 
v= 0.1176 mL 
m= 1.059 g/mL X 0.1176 mL= 0.1233624 g 

Now that we have the mass of acetic acid, we find the molar mass of acetic acid. 
M(acetic acid) = 60.05g/mol 

Then use the molar mass of acetic acid to find the amount of moles by diving the mass of acetic acid by its molar mass: 
0.1233626g / (60.05g/mol) = 0.002054321 mol 

Now based on stoichiometry, the ratio of vinegar to NaOH is 1:1 

All we have to do now is divide this number by the number of moles of NaOH and multiply by 1000 mL, to get the volume of vinegar in mL. 

so, (0.0020543 mol / 0.200 mol) * 1000mL = 10.3 mL 

The volume of 0.200 M NaOH that would be needed to titrate a 2 mL aliquot of this vinegar is 10.3 mL 

the answer is b. dilution