What mass of 95 % pure CaCO _ 3 will be required to neutralise 50mL...

(\underset{(s)}{\mathrm{CaCO}_3}+\underset{50 \mathrm{ mL}, 0.5 \mathrm{M}}{2 \mathrm{HCl}(\mathrm{aq})} \longrightarrow \underset{(\mathrm{aq})}{\mathrm{CaCl}_2}+\underset{(g)}{\mathrm{CO}_2}+\underset{(l)}{2 \mathrm{H}_2 \mathrm{O}}) the number of moles of HCl taken (=0.5 \times 0.05) $=0.025$ moles As we can see that from the above balanced equation is one mole of (\mathrm{CaCO}_3(s)) requires 2 moles of (\mathrm{HCl}(\mathrm{aq})) therefore, for 0.025 moles of (\mathrm{HCl}(\mathrm{aq}), 0.0125) moles of (\mathrm{CaCO}_3(s)) will be required. Mass of 0.0125 moles of (\mathrm{CaCO}_3(s)=0.0125 \times) molar mass of (\mathrm{CaCO}_3) $$\begin{aligned} & =0.0125 \times 100 \\ & =1.25 \mathrm{ g}\end{aligned}$$) But purity$of$(\mathrm{CaCO}_3(s)) is (95 \%). Therefore, the actual amount of (\mathrm{CaCO}_3(s)) required is Percentage purity (=\frac{\text{ weight of pure substance }}{\text{ weight of impure sample }} \times 100) (95=\frac{1.25}{\text{ weight of impure sample }} \times 100) weight of impure sample (=\frac{1.25 \times 100}{95}) (=1.32 \mathrm{ g}) Hence. option 1 is correct.$