100 mL of Na_3 PO_4 solution contains 3.45 g of sodium. The molarity of the solution is \ \ \ \ × 10^-2 mol L^-1. (Nearest integer) [Atomic Masses -…

$\displaystyle 100$ mL of $\displaystyle \mathrm{Na}_3 \mathrm{PO}_4$ solution contains $\displaystyle 3.45$ g of sodium. The molarity of the solution is \_\_\_\_ $\displaystyle \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$. (Nearest integer) [Atomic Masses - Na : $\displaystyle 23.0$ u, O : $\displaystyle 16.0$ u, P : $\displaystyle 31.0$ u]