Sodium oxide reacts with water to produce sodium hydroxide. 20.0 g of sodium oxide is dissolved in 500 mL of water. Neglecting the change in volume, the concentration of the resulting NaOH solution is _________ $$\times 10^{-1}$$ M. (Nearest integer)
[Atomic mass : Na = 23.0, O = 16.0, H = 1.0]

First, let us write the balanced chemical equation for the reaction between sodium oxide and water.

$$\mathrm{Na_2O + H_2O \;\longrightarrow\; 2\,NaOH}$$

This tells us that $$1$$ mole of $$\mathrm{Na_2O}$$ produces $$2$$ moles of $$\mathrm{NaOH}$$.

Now we calculate the molar mass of sodium oxide. The atomic masses supplied are $$\mathrm{Na = 23.0}$$ and $$\mathrm{O = 16.0}$$, so

Molar mass of $$\mathrm{Na_2O} \;=\; 2(23.0)\;+\;16.0 \;=\; 46.0\;+\;16.0 \;=\; 62.0\ \text{g mol}^{-1}.$$

We have been given $$20.0\ \text{g}$$ of $$\mathrm{Na_2O}$$. Using the definition

Moles $$= \frac{\text{Mass}}{\text{Molar mass}},$$

we get

$$n_{\mathrm{Na_2O}} = \frac{20.0\text{ g}}{62.0\text{ g mol}^{-1}} = 0.3226\text{ mol}\;(\text{keeping extra digits for accuracy}).

The stoichiometric ratio from the equation is $$1:2$$, so

$$n_{\mathrm{NaOH}} = 2 \times n_{\mathrm{Na_2O}} = 2 \times 0.3226 = 0.6452\ \text{mol}.$$

This amount of $$\mathrm{NaOH}$$ is dissolved in $$500\ \text{mL}$$ of water. Neglecting any volume change, the solution volume is

$$V = 500\ \text{mL} = 0.500\ \text{L}.$$

Molarity (concentration) is given by the formula

$$M = \frac{$$ Moles of solute $$}{$$ Volume of solution in litres $$}.$$

Substituting the values, we obtain

$$M = \frac{0.6452\text{ mol}}{0.500\text{ L}} = 1.2904\text{ M}.$$

We now express this value in the required form $$x \times 10^{-1}\ \text{M}$$. Writing $$1.2904\ \text{M}$$ as a multiple of $$10^{-1}$$, we have

$$1.2904\text{ M} = 12.904 \times 10^{-1}\text{ M}.$$

Rounding $$12.904$$ to the nearest integer gives $$13$$.

So, the answer is $$13$$.