In order to prepare $500\mathrm{ml}$ of $0.2\mathrm{M}$ $\mathrm{NaOH}$, the mass of $\mathrm{NaOH}$ required is

Given data

The volume of the solution is $500\mathrm{ml}$

The molarity of the solution is $0.2\mathrm{M}$, that is $0.2{\mathrm{molL}}^{-1}$

Step 1 Molar mass:

$\mathrm{Molarity}=\frac{\mathrm{Amount}\mathrm{of}\mathrm{solute}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{solute}}\times \frac{1000}{\mathrm{volume}\mathrm{of}\mathrm{solution}\left(\mathrm{ml}\right)}$

The atomic mass of $\mathrm{Na}$ is $23\mathrm{g}$

The atomic mass of $\mathrm{O}$ is $16\mathrm{g}$

The atomic mass of $\mathrm{H}$ is $1\mathrm{g}$

The molecular mass of $\mathrm{NaOH}$ is $23+16+1=40\mathrm{g}$

Step 2 Amount of Sodium hydroxide present:

Let the amount of solute be '$w$'

Therefore,

$\mathrm{Molarity}=\frac{\mathrm{Amount}\mathrm{of}\mathrm{solute}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{solute}}\times \frac{1000}{\mathrm{volume}\mathrm{of}\mathrm{solution}\left(\mathrm{ml}\right)}$

That is,

$$\begin{gathered} 0.2=\frac{\mathrm{w}}{40}\times \frac{1000}{500} \\ 0.2=\frac{\mathrm{w}}{40}\times 2 \\ 0.2=\frac{\mathrm{w}}{20} \\ \mathrm{w}=0.2\times 20 \\ \mathrm{w}=4\mathrm{g} \end{gathered}$$

Therefore, in order to prepare $500\mathrm{ml}$ of $0.2\mathrm{M}$ $\mathrm{NaOH}$, the mass of $\mathrm{NaOH}$ required is $4\mathrm{g}$.