A sodium hydroxide solution containing 40% by weight of pure $\mathrm{NaOH}$has a specific gravity $1.5$ of What volume (in m.L) of this solution required in the preparation of $500 \mathrm{mL}$ of a $0.45 M$$\mathrm{NaOH}$solution?

 $40\%= 0.066 \mathrm{Litre}$ means $40 gm NaOH$in 100 gram solution.  

Density of solution $= 1.5$

Volume of solution$=\frac{100}{1.5}=\frac{1000}{15}\left[\mathrm{D}=\frac{\mathrm{m}}{\mathrm{V}}\right]$

Volume of solution$= 66.66 ml$

Volume of solution

In 40 gm $\mathrm{NaOH}$, number of moles $= 40/\mathrm{m}. \mathrm{wt}=\frac{40}{40}=1$

Molarity of solution$=\frac{1}{0.066}=15\mathrm{M}$

$\begin{array}{l}{\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{M}}_2{\mathrm{V}}_2\\ 15{\mathrm{V}}_1=0.45\times 500\\ \mathrm{V}=15\mathrm{mL}\end{array}$