how many moles of NaOH (aq) is contained in 10.0mL of vinegar?
none. NaOH contains no acetic acid.
To determine the number of moles of NaOH (aq) present in 10.0 mL of vinegar, we need to make a few assumptions and perform some calculations.
Assumption:
Vinegar is a dilute acetic acid solution, and we assume it reacts completely with NaOH in a 1:1 molar ratio.
Steps to calculate the number of moles of NaOH:
1. Determine the concentration of NaOH solution:
- Look up the concentration of the NaOH solution. Let's assume it is 1.00 M (moles per liter).
2. Convert the volume of vinegar to liters:
- To work with the concentration in moles per liter (M), we need to convert the volume of vinegar from milliliters (mL) to liters (L).
- 1 L = 1000 mL, so 10.0 mL of vinegar is equal to 10.0/1000 = 0.010 L.
3. Calculate the number of moles of NaOH:
- We can use the equation: moles = concentration (M) × volume (L).
- The concentration of NaOH is 1.00 M, and the volume of vinegar is 0.010 L, so:
moles = 1.00 M × 0.010 L = 0.010 moles of NaOH.
Therefore, 10.0 mL of vinegar contains approximately 0.010 moles of NaOH (aq).