On heating `1.763 g` of hydrated `BaCl_2` to dryness, `1.505 g` of anhydrous salt remained. Hence the formula of the hydrate is (Atomic weight of`` Ba` = 137`)

Options

A. `BaCl_2 . 1/2 H_2O`
B. `BaCl_2 . H_2O`
C. `BaCl_2 . 2H_2O`
D. `BaCl_2 .5H_2O`

Answer

To determine the formula of the hydrated barium chloride (BaCl₂·xH₂O), we will follow these steps:

### Step 1: Calculate the mass of water lost
The mass of the hydrated barium chloride before heating is given as 1.763 g, and the mass of the anhydrous barium chloride remaining after heating is 1.505 g.

**Calculation:**
\[ \text{Mass of water lost} = \text{Mass of hydrated BaCl}_2 - \text{Mass of anhydrous BaCl}_2 \]
\[ \text{Mass of water lost} = 1.763 \, \text{g} - 1.505 \, \text{g} = 0.258 \, \text{g} \]

### Step 2: Calculate the moles of anhydrous BaCl₂
Next, we need to calculate the molar mass of anhydrous BaCl₂. The atomic weights are:
- Ba = 137 g/mol
- Cl = 35.5 g/mol (there are 2 Cl atoms)

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

Now, we can calculate the number of moles of anhydrous BaCl₂ using its mass.

**Calculation:**
\[ \text{Moles of anhydrous BaCl}_2 = \frac{\text{Mass of anhydrous BaCl}_2}{\text{Molar mass of BaCl}_2} \]
\[ \text{Moles of anhydrous BaCl}_2 = \frac{1.505 \, \text{g}}{208 \, \text{g/mol}} \approx 0.00724 \, \text{mol} \]

### Step 3: Calculate the moles of water lost
Next, we need to calculate the number of moles of water lost. The molar mass of water (H₂O) is 18 g/mol.

**Calculation:**
\[ \text{Moles of water lost} = \frac{\text{Mass of water lost}}{\text{Molar mass of water}} \]
\[ \text{Moles of water lost} = \frac{0.258 \, \text{g}}{18 \, \text{g/mol}} \approx 0.01433 \, \text{mol} \]

### Step 4: Determine the value of x in BaCl₂·xH₂O
To find the value of x, we can set up a ratio of the moles of water lost to the moles of anhydrous BaCl₂.

**Calculation:**
\[ \frac{\text{Moles of water}}{\text{Moles of anhydrous BaCl}_2} = \frac{0.01433}{0.00724} \approx 1.978 \approx 2 \]

Thus, x = 2.

### Conclusion
The formula of the hydrated barium chloride is BaCl₂·2H₂O.

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