0.7875 g of crystalline barium hydroxide is dissolved in water .For the neutralization of this solution 20 mL of N/4 `HNO_(3)` is required. How many moles of water of crystallization are present in one mole of this base ? (Given : Atomic mass Ba=137,O=16, N=14, H=1)

Answer

To solve the problem, we will follow a systematic approach to find the number of moles of water of crystallization in crystalline barium hydroxide (Ba(OH)₂·XH₂O).

### Step-by-Step Solution:

**Step 1: Determine the number of equivalents of HNO₃ used.**
- Given that 20 mL of N/4 HNO₃ is used for neutralization.
- Convert the volume from mL to L:
\[ \text{Volume in L} = \frac{20 \, \text{mL}}{1000} = 0.020 \, \text{L} \]
- Calculate the number of equivalents of HNO₃:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume in L} = \frac{1}{4} \times 0.020 = 0.005 \, \text{equivalents} \]

**Step 2: Set up the relationship between equivalents of acid and base.**
- For the neutralization reaction, the number of equivalents of the base (barium hydroxide) will equal the number of equivalents of the acid (HNO₃):
\[ \text{Equivalents of Ba(OH)₂·XH₂O} = 0.005 \]

**Step 3: Calculate the number of equivalents of barium hydroxide.**
- The n-factor for barium hydroxide (Ba(OH)₂) is 2 because it can donate 2 hydroxide ions (OH⁻).
- Therefore, the number of equivalents of Ba(OH)₂·XH₂O can be expressed as:
\[ \text{Equivalents} = \text{Number of moles} \times n \text{-factor} \]
\[ 0.005 = \text{Number of moles of Ba(OH)₂·XH₂O} \times 2 \]
\[ \text{Number of moles of Ba(OH)₂·XH₂O} = \frac{0.005}{2} = 0.0025 \, \text{moles} \]

**Step 4: Calculate the molar mass of barium hydroxide with water of crystallization.**
- The molar mass of barium hydroxide (Ba(OH)₂) is calculated as follows:
- Atomic mass of Ba = 137 g/mol
- Atomic mass of O = 16 g/mol (2 O in OH)
- Atomic mass of H = 1 g/mol (2 H in OH)
\[ \text{Molar mass of Ba(OH)₂} = 137 + (2 \times 16) + (2 \times 1) = 137 + 32 + 2 = 171 \, \text{g/mol} \]
- The molar mass of water (H₂O) is 18 g/mol. Therefore, the molar mass of barium hydroxide with X moles of water is:
\[ \text{Molar mass of Ba(OH)₂·XH₂O} = 171 + 18X \]

**Step 5: Use the mass of barium hydroxide to find X.**
- The mass of barium hydroxide solution is given as 0.7875 g:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.7875}{171 + 18X} \]
- Setting the number of moles equal to 0.0025:
\[ 0.0025 = \frac{0.7875}{171 + 18X} \]
- Cross-multiplying gives:
\[ 0.0025(171 + 18X) = 0.7875 \]
\[ 0.4275 + 0.045X = 0.7875 \]
\[ 0.045X = 0.7875 - 0.4275 = 0.36 \]
\[ X = \frac{0.36}{0.045} = 8 \]

**Step 6: Conclusion**
- The number of moles of water of crystallization present in one mole of barium hydroxide is **8**.