What mass of 95% pure CaCO_3 will be required to neutralise 50 mL of 0.5M HCl solution according to the following reaction? CaCO_3(s)+2 HCl_( aq)→…

$\underset{(s)}{\mathrm{CaCO}_3}+\underset{50 \mathrm{~mL}, 0.5 \mathrm{M}}{2 \mathrm{HCl}(\mathrm{aq})} \longrightarrow \underset{(\mathrm{aq})}{\mathrm{CaCl}_2}+\underset{(g)}{\mathrm{CO}_2}+\underset{(l)}{2 \mathrm{H}_2 \mathrm{O}}$(s)CaCO3​​+50 mL,0.5M2HCl(aq)​⟶(aq)CaCl2​​+(g)CO2​​+(l)2H2​O​ the number of moles of HCl taken $=0.5 \times 0.05$=0.5×0.05 $=0.025$=0.025 moles As we can see that from the above balanced equation is one mole of $\mathrm{CaCO}_3(s)$CaCO3​(s) requires 2 moles of $\mathrm{HCl}(\mathrm{aq})$HCl(aq) therefore, for 0.025 moles of $\mathrm{HCl}(\mathrm{aq}), 0.0125$HCl(aq),0.0125 moles of $\mathrm{CaCO}_3(s)$CaCO3​(s) will be required. Mass of 0.0125 moles of $\mathrm{CaCO}_3(s)=0.0125 \times$CaCO3​(s)=0.0125× molar mass of $\mathrm{CaCO}_3$CaCO3​ $\begin{aligned} & =0.0125 \times 100 \\ & =1.25 \mathrm{~g} \end{aligned}$​=0.0125×100=1.25 g​ But purity of $\mathrm{CaCO}_3(s)$CaCO3​(s) is $95 \%$95%. Therefore, the actual amount of $\mathrm{CaCO}_3(s)$CaCO3​(s) required is Percentage purity $=\frac{\text { weight of pure substance }}{\text { weight of impure sample }} \times 100$= weight of impure sample  weight of pure substance ​×100 $95=\frac{1.25}{\text { weight of impure sample }} \times 100$95= weight of impure sample 1.25​×100 weight of impure sample $=\frac{1.25 \times 100}{95}$=951.25×100​ $=1.32 \mathrm{~g}$=1.32 g Hence. option 1 is correct.