The density of 1.50 M solution of acetic acid (CH₃COOH) is 1.05 g cm ^-3. The molality of the solution will be

Let there be 1 L of solution. we have n₂ = 1.5 mol and V = 1 L

Mass of Solution $m=V\rho =\left({10}^3{\mathrm{cm}}^3\right)\left(1.05\mathrm{g}{\mathrm{cm}}^{-3}\right)=1050\mathrm{g}$

Mass of acetic acid, $m_2=n_2{M}_2=\left(1.5\mathrm{mol}\right)\left(60{\mathrm{gmol}}^{-1}\right)=90.0\mathrm{g}$

Mass of Water ,$m_1=m-m_2=1050\mathrm{g}-90\mathrm{g}=960\mathrm{g}=0.960\mathrm{kg}$

Molarity of solution, $m=\frac{n_2}{m_1}=\frac{1.5\mathrm{mol}}{0.960\mathrm{kg}}=1.56\mathrm{molkg}$