To solve this problem, we need to determine the mass of $95\%$ pure $CaCO_3$ required to neutralize $50$ mL of $0.5$ M $HCl. \\$The balanced chemical equation is:$\\ CaCO_3 (s) + 2HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + 2H_2O (l) \\$From the equation, 1 mole of $CaCO_3$ reacts with 2 moles of $HCl. \\$Step 1: Calculate the moles of $HCl$ present.$\\$Volume of $HCl = 50$ mL$= 0.050$ L$\\$Molarity of $HCl = 0.5$ M$\\$Moles of $HCl =$Molarity$\times$Volume$= 0.5 \times 0.050 = 0.025$ moles$\\$Step 2: Calculate the moles of $CaCO_3$ required.$\\$From the stoichiometry of the reaction,$\\ 1$ mole of $CaCO_3$ reacts with 2 moles of $HCl. \\$Moles of $CaCO_3 = \frac{Moles \ of \ HCl}{2} = \frac{0.025}{2} = 0.0125$ moles$\\$Step 3: Calculate the mass of pure $CaCO_3$ required.$\\$Molar mass of $CaCO_3 = 40.08 \, ($Ca$) + 12.01 \, ($C$) + 3 \times 16.00 \, ($O$) = 100.09 \,$g/mol$\\$Mass of pure $CaCO_3 =$Moles$\times$Molar mass$= 0.0125 \times 100.09 = 1.251125 \,$g$\\$Step 4: Calculate the mass of $95\%$ pure $CaCO_3$ required.$\\$Let $x$ be the mass of $95\%$ pure $CaCO_3. \\ 0.95x = 1.251125 \\ x = \frac{1.251125}{0.95} = 1.3175 \,$g$\\$Therefore, the mass of $95\%$ pure $CaCO_3$ required is approximately $1.32 \,$g.$\\$This corresponds to Option 2.