How much of $\mathrm{NaOH}$ is required to neutralize 1500 cm3 of 0.1 M HCl ?

$1500{\mathrm{cm}}^3$ of $0.1 \mathrm{NHCl}$ have number of gm equivalence $=\frac{N_1\times V_1}{1000}=\frac{1500\times 0.1}{1000}=0.15$

$\therefore 0.15\mathrm{gm}$. Equivalent of  $\mathrm{NaOH}=0.15\times 40=6\mathrm{gm}$.