1.00 L of 0.15 N NaOH absorbed 11.2 millimole of CO₂ from air. Hence, new molarity of NaOH is:

$\mathrm{NaOH}+{\mathrm{CO}}_2\to {\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O}$

1 L of 0.15 N

0.15 mol              0.0112 mol

NaOH required by 0.0112 mol CO₂ = 0.0224 mol
NaOH unreacted = 0.15 - 0.0224
Thus, M (NaOH) = 0.1276 M