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  4. How much of NaOH is required to neutralize 1500 cm3 of 0.1 N HCl?

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Q. How much of NaOH is required to neutralize 1500 $cm^3$ of 0.1 N HCl?

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Solution:

1500 $cm^3$ of 0.1 N HCl $\equiv$ 1500 $cm^3$ of 0.1 N NaOH
$\equiv$ 1500 $cm^3$ of 0.1 M NaOH
= $\left( \frac{1500}{1000} L \right) \times (0.1 \, mol \, L^{-1} )$ = 0.15 mol.
= 0.15 mol $\times$ 40 g mol$^{-1}$ = 6 g