posted by confused on .
a 10.0mL vinegar sample was completely neutralized by 22.5mL 0.2M NaOH solution. calculate molarity and % of acetic acid in vinegar.
I assume we call the density of the vinegar 1.00 g/mL so the 10 mL has a mass of 10.0 grams.
moles NaOH used = M x L = ??
moles of acetic acid in the 10.0 mL = ?? (same as moles NaOH).
M acetic acid = moles/L = ??/0.1 = xx M.
Is that percent w/w or w/v. I assume w/v.
You have M and that is moles/L.
moles x molar mass = grams so that is g/L
You want g/100 mL = ??
i'm still lost can you show me step by step please
I worked the problem for you. All you need to do is substitute a few numbers. Post your work and tell me what you don't understand at the next step and I'll try to help you through BUT I am not going to do the problem for you.
is the answer M= .08M
and is the % of mass= 44%