Molarity of 1m aqueous NaOH solution [density of the solution is 1.02 g/ml]

$\therefore \mathrm{m}=\frac{\mathrm{M}\times 1000}{(1000\times \mathrm{d})-{\mathrm{MM}}^1}$

M = molarity of the solution
m = molality of the solution
d = density of the solution (in g. ml^-1)
M¹= gram molecular weight of solute

$$\begin{gathered} 1\mathrm{m}=\frac{\mathrm{M}\times 1000}{(1000\times 1.02)-\mathrm{M}\left(40\right)} \\ \mathrm{M}=0.98 \end{gathered}$$