10 g of a hydrated salt of `BaCI_2` was dissolved in one litre water. The solution was then treated with 1.65 litre of 0.05 N `AgNO_3` till complete precipitation. The number of water molecules in hydrated salt are :

Options

A. 6
B. 10
C. 4
D. 2

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Define the hydrated salt
Let the hydrated salt be represented as \( \text{BaCl}_2 \cdot n \text{H}_2\text{O} \), where \( n \) is the number of water molecules.

### Step 2: Calculate the molar mass of the hydrated salt
The molar mass of \( \text{BaCl}_2 \) is calculated as follows:
- Molar mass of Ba = 137 g/mol
- Molar mass of Cl = 35.5 g/mol (2 Cl atoms)

\[ \text{Molar mass of BaCl}_2 = 137 + (2 \times 35.5) = 137 + 71 = 208 \text{ g/mol} \]

The molar mass of the hydrated salt \( \text{BaCl}_2 \cdot n \text{H}_2\text{O} \) is:
\[ \text{Molar mass} = 208 + 18n \]

### Step 3: Calculate the equivalents of \( \text{AgNO}_3 \)
Given:
- Volume of \( \text{AgNO}_3 = 1.65 \) L
- Normality of \( \text{AgNO}_3 = 0.05 \) N

The number of equivalents of \( \text{AgNO}_3 \) can be calculated using the formula:
\[ \text{Equivalents} = \text{Normality} \times \text{Volume} \]
\[ \text{Equivalents of AgNO}_3 = 0.05 \times 1.65 = 0.0825 \text{ equivalents} \]

### Step 4: Calculate the equivalents of \( \text{BaCl}_2 \cdot n \text{H}_2\text{O} \)
The n-factor for \( \text{BaCl}_2 \) is 2 because it can produce 2 moles of \( \text{Cl}^- \) ions.

Using the formula for equivalents:
\[ \text{Equivalents of BaCl}_2 \cdot n \text{H}_2\text{O} = n \times \text{Molarity} \times \text{Volume} \]

Since the solution is 1 L, we can express the molarity as:
\[ \text{Molarity} = \frac{\text{mass}}{\text{molar mass} \times \text{volume}} = \frac{10}{(208 + 18n) \times 1} \]

Thus, the equivalents of \( \text{BaCl}_2 \cdot n \text{H}_2\text{O} \) can be expressed as:
\[ 2 \times \frac{10}{(208 + 18n)} = 0.0825 \]

### Step 5: Set up the equation
Now we can set up the equation:
\[ \frac{20}{208 + 18n} = 0.0825 \]

### Step 6: Solve for \( n \)
Cross-multiplying gives:
\[ 20 = 0.0825 \times (208 + 18n) \]
\[ 20 = 17.22 + 1.485n \]
\[ 20 - 17.22 = 1.485n \]
\[ 2.78 = 1.485n \]
\[ n = \frac{2.78}{1.485} \approx 1.87 \approx 2 \]

### Conclusion
The number of water molecules in the hydrated salt \( \text{BaCl}_2 \cdot n \text{H}_2\text{O} \) is approximately 2.