$\mathrm{150~mL}$ of $\mathrm{0.1~N}$ $\ce{HCl}$ is required to react with $\mathrm{1~g}$ of a sample of limestone. Calculate percentage purity of calcium carbonate.
Reaction: $$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$
Number of milli equivalence of $\ce{HCl}$ used initially is $\mathrm{15}$
Number of milli moles of $\ce{HCl}$ used initially is $\mathrm{\frac{15}{2}}$
$\mathrm{2}$ moles of $\ce{HCl}$ reacts with $\mathrm{1}$ mole of $\ce{CaCO3}$.
So amount of calcium carbonate is $\mathrm{\frac{15}{4}\times 10^{-3}~mol}=\mathrm{\frac{3}{8}~g}$
But answer given is $\mathrm{75\%}$