A 100 mL solution was made by adding 1.43 g of $$Na_2CO_3 \cdot xH_2O$$. The normality of the solution is 0.1 N. The value of x is __________ (The atomic mass of Na is 23 g/mol)

We start by denoting the unknown number of water molecules in the hydrated salt as $$x$$ in $$Na_2CO_3\cdot xH_2O$$. Our aim is to relate the given mass of the salt to the normality of the prepared solution and thereby determine $$x$$.

The first quantity given is the normality of the solution: $$N = 0.1\ \text{N}$$. Normality is defined as

$$N = \dfrac{\text{Number of gram-equivalents of solute}}{\text{Volume of solution in litres}}.$$

For any salt that reacts with acids or bases, the number of gram-equivalents is obtained from

$$\text{Gram-equivalents} = n\text{ (moles)} \times \text{n-factor},$$

where the n-factor is the total number of replaceable $$\text{H}^+$$ or $$\text{OH}^-$$ ions (or the total charge transferred) per mole of the substance in the relevant reaction. In the usual acid-base context, one mole of $$Na_2CO_3$$ can accept two protons, so its n-factor is $$2$$. Hence,

$$N = M \times \text{n-factor},$$

where $$M$$ is the molarity. Substituting $$\text{n-factor} = 2$$, we get

$$M = \dfrac{N}{2} = \dfrac{0.1}{2} = 0.05\ \text{M}.$$

Now, the volume of the prepared solution is $$100\ \text{mL} = 0.1\ \text{L}$$. The number of moles of the hydrated salt present is therefore

$$n = M \times V = 0.05\ \text{mol\,L}^{-1} \times 0.1\ \text{L} = 0.005\ \text{mol}.$$

Next we calculate the molar mass of the hydrated salt $$Na_2CO_3\cdot xH_2O$$. The anhydrous part $$Na_2CO_3$$ has the molar mass

$$\begin{aligned} M_{Na_2CO_3} &= 2(23)\;+\;12\;+\;3(16)\\ &= 46\;+\;12\;+\;48\\ &= 106\ \text{g\,mol}^{-1}. \end{aligned}$$

Each water molecule contributes $$18\ \text{g\,mol}^{-1}$$, so the total molar mass of the hydrate is

$$M_{\text{hydrate}} = 106 + 18x\ \text{g\,mol}^{-1}.$$

The sample mass given is $$1.43\ \text{g}$$. Using the relation

$$n = \dfrac{\text{mass}}{\text{molar mass}},$$

we write

$$0.005 = \dfrac{1.43}{106 + 18x}.$$

Now we cross-multiply to isolate $$x$$:

$$1.43 = 0.005\,(106 + 18x).$$

Dividing both sides by $$0.005$$ gives

$$106 + 18x = \dfrac{1.43}{0.005} = 286.$$

Subtracting $$106$$ from both sides, we obtain

$$18x = 286 - 106 = 180.$$

Finally, dividing by $$18$$, we find

$$x = \dfrac{180}{18} = 10.$$

So, the answer is $$10$$.