2.86 g of Na₂CO₃.xH₂O in 100 ml solution is 0.2 N. Then the value of x is,

$\begin{array}{l}{\text{molar mass of Na}}_{2}c{o}_3.x{H}_{2}o=(106+18x)\\ 2.86\text{ g salt = }\left(\frac{2.86}{106+18x}\right)\text{ mole in 100mL}\\ \text{ Molarity = }\left(\frac{28.6}{106+18x}\right)\text{ mole in 1000mL}\\ For{\text{ Na}}_{2}C{o}_3\text{, normality=2Xmolarity=0.2N}\\ \left(\frac{2\text{X}28.6}{106+18x}\right)=0.2\\ \text{ x}=10\end{array}$