Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25°C. Calculate the molarity of a solution of acetic acid made by dissolving 35.00 mL of glacial acetic acid at 25°C in enough water to make 400.0 mL of solution.

Answer 1

Gms of HOAc = (35.0ml)(1.049g/ml) = 36.715gms HOAc
moles HOAc = (36.715 g)/(60.05 g/mol) = 0.6114 mole
Molarity = moles/Liters
= (0.6114-mole/0.400-L)
=1.53-Molar

Answer 2

First determine the M of the glacial acetic acid.
1000 mL x 1.049 g/mL x 1.00 x (1/molar mass acetic acid) = ?M