A sample of crystalline `Ba(OH)_(2). XH_(2)O` weight 1.578 gm was dissolved in water. The solution required 40 ml of `0.25 N HNO_(3)` for complete relation. Determine the number of molecular of water of crystallisat in base.

Answer

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**.