The mass of sodium carbonate $\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)$ to prepare $400\mathrm{mL}$ of $0.275$ molar aqueous solution is 

Since molarity,$M=n_2/V$, we have 

Amount of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ required,$n_2=MV=\left(0.275{\mathrm{molL}}^{-1}\right)\left(0.40\mathrm{L}\right)=0.11\mathrm{mol}$

Mass of  ${\mathrm{Na}}_2{\mathrm{CO}}_3$ required, $m=n_2{M}_2=\left(0.11\mathrm{mol}\right)\left(106\mathrm{g}{\mathrm{mol}}^{-1}\right)=11.66\mathrm{g}$.