The density of 2 M aqueous solution of NaOH is 1.28 ~g / cm^3 . The...

Given, molarity $=2 \mathrm{M}$ [i.e 2 mole $\mathrm{NaOH}$ in $1 \mathrm{~L}$ solution] Density $(\mathrm{d})=1.28 \mathrm{~cm}^{-3}$ Molecular weight of $\mathrm{NaOH}=40 \mathrm{gmol}^{-1}$ We know that, $$\begin{aligned}& \text{ Density }=\frac{\text{ mass of solution }}{\text{ volume of solution }}\\ & \therefore \text{ Mass of solution }=1.28 \mathrm{~g} \mathrm{~cm}^{-3} \times1000 \mathrm{~mL}=1280 \mathrm{~g}\\ & \text{ Moreover, molarity }\\ & =\frac{\text{ number of moles of solute }}{\text{ volume of solution }(\text{ in } \mathrm{mL})} \times1000 \\ & \therefore \quad 2=\frac{\text{ mass of solute }}{40 \times 1000} \times1000=80 \mathrm{~g}\\ & \text{ Now, mass of solution }=\text{ mass of solvent }+ \text{ mass of }\\ & \text{ solute }\\ & 1280 \mathrm{~g}=\text{ mass of solvent }+80 \\ & \therefore\text{ Mass of solvent }=1280-80=1200 \mathrm{~g}=1.2 \mathrm{~kg}\\ & \text{ Now, molality }=\frac{\text{ number of moles of solute }}{\text{ mass of solvent }(\text{ in kg) }}\\ & =\frac{2}{1.2}=\frac{20}{12}=\frac{5}{3}=1.67 \mathrm{~m}\\ & \end{aligned}$$Alternative method Molality $(\mathrm{m})=\frac{\mathrm{M}}{1000 \mathrm{~d}-\mathrm{M} \omega \mathrm{t}} \times 1000$ where, $M=$ molarity, $d=$ density of solution $\mathrm{M} \omega \mathrm{t}=\text{ molar mass of solute. }On substing the given values, we get$$\begin{aligned}\text{ Molality }& =\frac{2}{1000 \times 1.28-40} \times1000 \\ & =\frac{2}{1280-40} \times1000=\frac{2000}{1240}=1.612 \mathrm{~m} \end{aligned}$$