To solve this problem, we need to determine how much sodium hydroxide (NaOH) remains unreacted after reacting with hydrochloric acid (HCl).$\\$Given:$\\$• Mass of NaOH $= 1$ gram$\\$• Volume of HCl solution $= 25$ mL$\\$• Molarity of HCl solution $= 0.75$ M$\\$Step 1: Calculate moles of HCl.$\\$Moles of HCl$= \frac{Molarity \times \ Volume \ in \ L}{1} \\ = 0.75 \times \frac{25}{1000} \\ = 0.01875$ moles$\\$Step 2: Calculate moles of NaOH.$\\$Molar mass of NaOH$= 23 + 16 + 1 = 40$ g/mol$\\$Moles of NaOH$= \frac{Mass}{Molar \ mass} \\ = \frac{1}{40} \\ = 0.025$ moles$\\$Step 3: Determine the reaction.$\\$The balanced chemical equation is:$\\$NaOH$+$HCl$\rightarrow$NaCl$+$H$_2$O$\\$This shows a 1:1 mole ratio between NaOH and HCl.$\\$Step 4: Calculate moles of NaOH left unreacted.$\\$Initially, we have 0.025 moles of NaOH and 0.01875 moles of HCl.$\\$Since the reaction is 1:1, 0.01875 moles of NaOH will react with 0.01875 moles of HCl.$\\$Moles of NaOH left unreacted$= 0.025 - 0.01875 \\ = 0.00625$ moles$\\$Step 5: Calculate mass of unreacted NaOH.$\\$Mass of unreacted NaOH$=$Moles$\times$Molar mass$\\ = 0.00625 \times 40 \\ = 0.25$ grams$\\ = 250$ mg$\\$Therefore, the mass of sodium hydroxide left unreacted is 250 mg.$\\$This corresponds to Option 4.