How much of NaOH is required to Neutralize $1500 {\mathrm{cm}}^3$ of 0.1 N HCl ?

Normality of HCl solution = 0.1N

Volume of HCl solution, $\mathrm{V}=1500 {\mathrm{cm}}^3$

Since $1{\mathrm{cm}}^3=1\mathrm{mL}$

GMW of NaOH = 40 g/mol

GEW of NaOH$=\frac{\mathrm{GMW} \mathrm{of} \mathrm{NaOH}}{\mathrm{Acidity} \mathrm{of} \mathrm{NaOH}}=\frac{40}{1}=40$

The formula for normality of the solution is,

$\mathrm{N}=\frac{\mathrm{W}}{\mathrm{GEW}}\times \frac{1000}{\mathrm{V} \mathrm{in} \mathrm{mL}}$

$0.1=\frac{\mathrm{w}}{40}\times \frac{10 \cancel{00}}{15\cancel{00}}$

$\therefore \mathrm{w}=\frac{0.10\times 40\times 15}{10}=6.0\mathrm{g}$