The molarity of the solution prepared by dissolving 6.3 g of oxalic acid ( H_2 C_2 O_4 · 2 H_2 O ) in 250 mL of water in mol L^-1 is x × 10^-2. The…

The molarity of the solution prepared by dissolving $\displaystyle 6.3$ g of oxalic acid $\displaystyle \left(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}\right)$ in $\displaystyle 250$ mL of water in $\displaystyle \mathrm{mol} \mathrm{L}^{-1}$ is $\displaystyle x \times 10^{-2}$. The value of $\displaystyle x$ is \_\_\_\_. (Nearest integer) [Atomic mass : H : $\displaystyle 1.0$, C : $\displaystyle 12.0$, O : $\displaystyle 16.0$ ]