14.3 g of $N{a}_{2}C{O}_3.x{H}_{2}O$ completely neutralizes 100 ml of 1N $H_{2}S{O}_4$ solutions. Value of x is

No. of gram equivalent of the $N{a}_{2}C{O}_3.x{H}_{2}O$ = no of gram equivalent of $H_{2}S{O}_4$

gram equivalent of the $N{a}_{2}C{O}_3.x{H}_{2}O$:

 $$\begin{gathered} N{a}_{2}C{O}_3.x{H}_{2}O=\frac{given mass}{equivalent mass} \\ N{a}_{2}C{O}_3.x{H}_{2}O=\frac{14.3}{{\displaystyle \frac{106+18x}{2}}\left({\displaystyle \frac{molecular mass}{2}}=equivalent mass\right)} \end{gathered}$$...(1)

gram equivalent of $H_{2}S{O}_4=normality\times volume \left(L\right)=1\times 100\times {10}^{-3}...\left(2\right)$

Equate (1) and (2)

$$\begin{gathered} \frac{14.3}{{\displaystyle \frac{106+18x}{2}}}=1\times 100\times {10}^{-3} \\ x=10 \end{gathered}$$