Question:
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 $x$ is ___________ .
(The atomic mass of $\mathrm{Na}$ is $23 \mathrm{~g} / \mathrm{mol}$ )
Solution:
(10)
Normality $=\frac{\text { No. of equivalents of solute }}{\text { Volume of solution (in L) }}$
$0.1=\frac{1.43}{\frac{(106+18 x)}{2} \times 0.1} \Rightarrow \frac{106+18 x}{2}=143$
$\Rightarrow 18 x=286-106=180 \Rightarrow x=10$