Volume of $3 \mathrm{M} \mathrm{~NaOH}$ (formula weight $40 \mathrm{~g} \mathrm{~mol}^{-1}$ ) which can be prepared from $84 \mathrm{~g}$ of $\mathrm{NaOH}$ is __________ $\times 10^{-1} \mathrm{dm}^3$.