Answer
0.0408L.
Work Step by Step
1. Calculate the molar mass $(CH_3COOH)$:
12.01* 1 + 1.008* 3 + 12.01* 1 + 16.00* 2 + 1.008* 1 = 60.05g/mol
2. According to the balance coefficients, the ratio of acetic acid and $KOH$ moles is 1 to 1.
The molarity of the potassium hydroxide solution is 0.157 M.
3. Use thia information as conversion factors to find the required volume of that $KOH$ solution to react with 0.385 g of acetic acid:
$0.385g (CH_3COOH) \times \frac{1mol(CH_3COOH)}{60.05g(CH_3COOH)} \times \frac{1mol(KOH)}{1mol(CH_3COOH)} \times \frac{1L(KOH-solution)}{0.157mol(KOH)} = 0.0408L (KOH-solution)$
