How many grams of $NaOH$ will be required to neutralize 12.2 g benzoic acid

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+\mathrm{NaOH}⟶{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}$

For complete neutralization,

Mole of $NaOH$ = mole of benzoic acid

$\frac{w}{40}=\frac{12.2}{122}\Rightarrow w=4\mathrm{g}$.