The density of 40% by mass of acetic acid is 1.25 g mL^-1 • The molarity of the solution is

40 mass percent of acetic acid implies that

Mass of acetic acid, m₂ = 40 g; Mass of water, m ₁ = 60 g

Amount of acetic acid, $n_2=\frac{m_2}{M_2}=\frac{40\mathrm{g}}{60\mathrm{g}{\mathrm{mol}}^{-1}}=(2/3)\mathrm{mol}$

Volume of solution, $V=\frac{m_1+m_2}{\rho }=\frac{100\mathrm{g}}{1.25\mathrm{g}{\mathrm{mL}}^{-1}}=80\mathrm{mL}$

Molarity of solution, $M=\frac{n_2}{V}=\frac{(2/3)\mathrm{mol}}{80\times {10}^{-3}\mathrm{L}}=8.33\mathrm{M}$