The mass of acetic acid contained in 0.5 l of CH3COOH solution with a mass fraction of 80% (density 1.1 g / ml) is equal to?

Given:
V solution (CH3COOH) = 0.5 l = 500 ml
ω (CH3COOH) = 80%
ρ solution (CH3COOH) = 1.1 g / ml

To find:
m (CH3COOH) -?

Decision:
1) Calculate the mass of the acetic acid solution:
m solution (CH3COOH) = ρ solution (CH3COOH) * V solution (CH3COOH) = 1.1 * 500 = 550 g;
2) Calculate the mass of acetic acid in solution:
m (CH3COOH) = ω (CH3COOH) * m solution (CH3COOH) / 100% = 80% * 550/100% = 440 g.

Answer: The mass of acetic acid in solution is 440 g.