Explanation:

(b) \(\mathrm{NaOH}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}\)
Thus, 1 mole of NaOH reacts with 1 mole of HCl .
Now, Number of moles present are
\(\mathrm{NaOH}=\frac{1 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}=0.025 \mathrm{~mol}\)
\(\mathrm{HCl}=0.025 \mathrm{~L} \times 0.75 \mathrm{M}=0.01875 \mathrm{~mol}\).
\(\Rightarrow\) Number of moles of NaOH left \(=\) Number
of moles of NaOH present - Number of moles of HCl present (limiting reagent)
\(=0.025-0.01875=0.00625 \mathrm{~mol}\)
\(\Rightarrow\) Mass of NaOH left \(=0.00625 \times 40\)
\(=0.25 \mathrm{~g}=250 \mathrm{mg}\)