What mass of $95\%$ pure ${\mathrm{CaCO}}_3$ will be required to neutralise $50 \mathrm{mL}$ of $0.5 \mathrm{M} \mathrm{HCl}$ solution according to the following reaction?

${\mathrm{CaCO}}_{3\left(\mathrm{s}\right)}+2{\mathrm{HCl}}_{\left(\mathrm{aq}\right)}\to {\mathrm{CaCl}}_{2\left(\mathrm{aq}\right)}+{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

[Calculate upto second place of decimal point]