Howmany mL of .25MNaOH is needed to titrate 5.00mLsampleof vinegar thats4% by mass acetic acid &density1.0g/mL

Answer 1

4% by mass means that you have 4 gr CH3COOH per 100 gr of solution. However since d=1.0 g/ml it also means that the 100 gr of solution are equal to 100 ml.
The molecular weight is MW=60
So the Macid= mass/(MW*V)= 4/(60*0.1)= 2/3

CH3COOH + NaOH -> CH3COONa + H2O

the stoichiometry of the reaction is 1:1

Thus mole CH3COOH= mole NaOH =>
Macid *Vacid= Mbase* Vbase =>

Vbase= (Macid/Mbase)* Vacid=
=(2/3)* (1/0.25)* 5.00= 13.33 ml
Since the volume of acid is given with 2 decimal digits you should provide the volume of the base with the same number of decimal digits.

Answer 2

NaOH + Acetic Acid --------> H20 + Sodium Acetate

4gm Acetic acid/(0.1 Liter) *(1 mole Acetic acid/60.052g) = 0.666M Acetic Acid Solution.

(0.666 Mole Acetic acid/1000ml) * 5 ml = (0.25 Mole NaOH/1000ml)*(x ml)

solve for x, x = 13.322, round to 13.3 ml of 0.25 M NaOH

Please check for errors