A 100 mL solution was made by adding 1.43 g of


A $100 \mathrm{~mL}$ solution was made by adding $1.43 \mathrm{~g}$ of $\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot x \mathrm{H}_{2} \mathrm{O}$. The normality of the solution is $0.1 \mathrm{~N}$. The value of $x$ isĀ  ___________ .

(The atomic mass of $\mathrm{Na}$ is $23 \mathrm{~g} / \mathrm{mol}$ )



Normality $=\frac{\text { No. of equivalents of solute }}{\text { Volume of solution (in L) }}$

$0.1=\frac{1.43}{\frac{(106+18 x)}{2} \times 0.1} \Rightarrow \frac{106+18 x}{2}=143$

$\Rightarrow 18 x=286-106=180 \Rightarrow x=10$

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