The density of 1.40 molal solution of acetic acid (CH₃COOH) is 1.084 g mL^-1 • The molarity of solution will be

Let the mass of solvent be 1 kg. We will have m₁ = 1000 g and n₂ = 1.40 mol

Mass of acetic acid, $m_2=n_2{M}_2=\left(1.40\mathrm{mol}\right)\left(60\mathrm{g}{\mathrm{mol}}^{-1}\right)=8.40\mathrm{g}$

Mass of solution, $m=m_1+m_2=1000\mathrm{g}+84\mathrm{g}=1084\mathrm{g}$

Volume of solution, $V=\frac{m}{\rho }=\frac{1084\mathrm{g}}{1.084\mathrm{g}{\mathrm{mL}}^{-1}}=1000\mathrm{mL}=1\mathrm{L}$

Molarity of solution, $M=\frac{n_2}{V}=\frac{1.40\mathrm{mol}}{1\mathrm{L}}=1.40\mathrm{M}$