The mass of $N{a}_{2}C{O}_3$ (91.2% purity) required for the neutralisation of 45.6 mL of 0.25 M $\mathrm{HCl}$ solution is

The reaction is ${\mathrm{Na}}_2{\mathrm{CO}}_3+2\mathrm{HCl}\to 2\mathrm{NaCl}+{\mathrm{H}}_2{\mathrm{CO}}_3$

Amount of $\mathrm{HCl}$ to be neutralised $=VM=\left(45.6\times {10}^{-3}\mathrm{L}\right)\left(0.25{\mathrm{molL}}^{-1}\right)=11.4\times {10}^{-3}\mathrm{mol}$

Amount of $N{a}_{2}C{O}_3$ required $=0.5\times 11.4\times {10}^{-3}\mathrm{mol}=5.7\times {10}^{-3}\mathrm{mol}$

Mass of $N{a}_{2}C{O}_3$ required $=\left(5.7\times {10}^{-3}\mathrm{mol}\right)\left(106\mathrm{g}{\mathrm{mol}}^{-1}\right)(100/91.2)=0.6625\mathrm{g}$