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]
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In reaction-based concentration problems, always check the stoichiometry first. Here, the factor of 2 (1 mole $\text{Na}_2\text{O} \to$ 2 moles $\text{NaOH}$) is the most critical step often missed by students.
Step 1: Understanding the Concept:
Sodium oxide ($\text{Na}_2\text{O}$) is a basic oxide that reacts completely with water to form sodium hydroxide ($\text{NaOH}$). The concentration (molarity) of the solution is determined by the total moles of $\text{NaOH}$ produced and the final volume of the solution. Step 2: Key Formula or Approach:
1. Reaction equation: $\text{Na}_2\text{O} + \text{H}_2\text{O} \to 2\text{NaOH}$
2. Molarity ($M$) = $\frac{\text{Moles of solute}}{\text{Volume of solution in L}}$
3. Molar mass of $\text{Na}_2\text{O} = 2 \times 23.0 + 16.0 = 62.0 \text{ g mol}^{-1}$. Step 3: Detailed Explanation:
1. Calculate the moles of $\text{Na}_2\text{O}$ dissolved:
\[ n(\text{Na}_2\text{O}) = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{20.0 \text{ g}}{62.0 \text{ g mol}^{-1}} \approx 0.32258 \text{ mol} \]
2. From the stoichiometry of the reaction, 1 mole of $\text{Na}_2\text{O}$ produces 2 moles of $\text{NaOH}$:
\[ n(\text{NaOH}) = 2 \times n(\text{Na}_2\text{O}) = 2 \times 0.32258 = 0.64516 \text{ mol} \]
3. Volume of solution $V = 500 \text{ mL} = 0.5 \text{ L}$.
4. Calculate Molarity ($M$):
\[ M = \frac{0.64516 \text{ mol}}{0.5 \text{ L}} = 1.29032 \text{ M} \]
5. Expressing in terms of $x \times 10^{-1}$:
\[ 1.29032 = 12.9032 \times 10^{-1} \]
The nearest integer for $x$ is 13. Step 4: Final Answer:
The value of $x$ is 13.
