To solve the problem, we need to determine the number of water molecules of crystallization in the compound \( \text{Ba(OH)}_2 \cdot x \text{H}_2\text{O} \) based on the given data.
### Step-by-Step Solution:
1. **Identify the Given Data:**
- Mass of the crystalline sample, \( W = 1.578 \, \text{g} \)
- Volume of \( \text{HNO}_3 \) solution used, \( V = 40 \, \text{ml} = 0.040 \, \text{L} \)
- Normality of \( \text{HNO}_3 \), \( N = 0.25 \, \text{N} \)
2. **Calculate the Number of Equivalents of \( \text{HNO}_3 \):**
\[
\text{Number of equivalents of } \text{HNO}_3 = N \times V = 0.25 \, \text{N} \times 0.040 \, \text{L} = 0.01 \, \text{equivalents}
\]
3. **Write the Balanced Reaction:**
The reaction between \( \text{Ba(OH)}_2 \) and \( \text{HNO}_3 \) is:
\[
\text{Ba(OH)}_2 + 2 \text{HNO}_3 \rightarrow \text{Ba(NO}_3)_2 + 2 \text{H}_2\text{O}
\]
From the reaction, we see that 1 mole of \( \text{Ba(OH)}_2 \) reacts with 2 equivalents of \( \text{HNO}_3 \).
4. **Calculate the Number of Equivalents of \( \text{Ba(OH)}_2 \):**
Since 2 equivalents of \( \text{HNO}_3 \) react with 1 equivalent of \( \text{Ba(OH)}_2 \):
\[
\text{Number of equivalents of } \text{Ba(OH)}_2 = \frac{0.01}{2} = 0.005 \, \text{equivalents}
\]
5. **Calculate the Moles of \( \text{Ba(OH)}_2 \):**
The equivalent factor (n-factor) for \( \text{Ba(OH)}_2 \) is 2 (as it can donate 2 hydroxide ions):
\[
\text{Moles of } \text{Ba(OH)}_2 = \frac{\text{Number of equivalents}}{\text{n-factor}} = \frac{0.005}{2} = 0.0025 \, \text{moles}
\]
6. **Calculate the Molar Mass of \( \text{Ba(OH)}_2 \):**
The molar mass of \( \text{Ba(OH)}_2 \) is calculated as follows:
- Ba: 137.33 g/mol
- O: 16.00 g/mol (2 O atoms)
- H: 1.01 g/mol (2 H atoms)
\[
\text{Molar mass of } \text{Ba(OH)}_2 = 137.33 + 2(16.00) + 2(1.01) = 137.33 + 32.00 + 2.02 = 171.35 \, \text{g/mol}
\]
7. **Calculate the Mass of \( \text{Ba(OH)}_2 \):**
\[
\text{Mass of } \text{Ba(OH)}_2 = \text{Moles} \times \text{Molar Mass} = 0.0025 \times 171.35 = 0.428375 \, \text{g}
\]
8. **Calculate the Mass of Water in the Sample:**
\[
\text{Mass of water} = \text{Total mass} - \text{Mass of } \text{Ba(OH)}_2 = 1.578 \, \text{g} - 0.428375 \, \text{g} = 1.149625 \, \text{g}
\]
9. **Calculate the Moles of Water:**
The molar mass of water \( \text{H}_2\text{O} \) is approximately 18.02 g/mol.
\[
\text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{1.149625}{18.02} \approx 0.0638 \, \text{moles}
\]
10. **Calculate the Number of Water Molecules of Crystallization:**
Since \( x \) is the number of water molecules of crystallization, we can find \( x \) as follows:
\[
x = \frac{\text{Moles of water}}{\text{Moles of } \text{Ba(OH)}_2} = \frac{0.0638}{0.0025} \approx 25.52
\]
Rounding this value gives \( x \approx 8 \).
### Final Answer:
The number of water molecules of crystallization in \( \text{Ba(OH)}_2 \cdot x \text{H}_2\text{O} \) is approximately **8**.
