Calculate the molarity of 5% acetic acid

given:
[CH3COOH] = 5%
Volume of acetic acid = 10.00ml

How do you find out the molarity of the acid?

If you assume the density of the solution is 1.00 g/mL, then mass HC2H3O2 = 0.05 x 10. mL = 0.5 g
? M HC2H3O2 = 0.5 g HC2H3O2 / 60 g/mol HC2H3O2 / 0.010 L = 0.83 M HC2H3O2
This is approximate since the density is more than 1 g/mL.

(5g/100g) x (1.049g/1mL) x (10.00mL/1L) x (1mol/60.05g) = 8.73x10^-3 M of acetic acid.

note: 1.049g/1mL is the density.