The density of 3 molal solution of NaOH is 1.110 gm/ml. Molarity of the solution is

Mass of solvent is calculated as:

$$\begin{gathered} \mathrm{Molality}=\frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute} \left(\mathrm{n}\right)}{\mathrm{Mass} \mathrm{of} \mathrm{solvent} (\mathrm{in} \mathrm{kg})} \\ 3=\frac{\mathrm{n}}{\mathrm{Mass} \mathrm{of} \mathrm{solvent} (\mathrm{in} \mathrm{kg})} \\ \mathrm{Mass} \mathrm{of} \mathrm{solvent} (\mathrm{in} \mathrm{g})=\frac{1000\times \mathrm{n}}{3} \mathrm{g} \end{gathered}$$

Number of moles is determined by dividing mass and molar mass. Moles of solute (NaOH) is calculated as:

$$\begin{gathered} \mathrm{Moles}=\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{mass}} \\ \mathrm{Moles}=\frac{\mathrm{m}}{40 \mathrm{g}/\mathrm{mol} (\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{NaOH})} \end{gathered}$$

Density is calculated by dividing the mass of solution (mass of solvent and mass of solute) with the volume of the solution.

$$\begin{gathered} \frac{\mathrm{Mass} \mathrm{of} \mathrm{solution}}{\mathrm{Volume}}=1.11 \\ \frac{\mathrm{Mass} \mathrm{of} \mathrm{solute}+\mathrm{Mass} \mathrm{of} \mathrm{solvent}}{\mathrm{Volume}}=1.11 \\ \frac{\mathrm{n}\times 40+1000\times \mathrm{n}}{3}=1.11 \\ \mathrm{n}=2.97\times {10}^{-3} \mathrm{mol} \end{gathered}$$

Molarity of NaOH is calculated as:

$\mathrm{Molarity}=\frac{2.97\times {10}^{-3} \mathrm{mol}}{0.001}=2.97 \mathrm{M}$