What is the percentage loss in water when `BaCI_2. 2H_2 O` becomes completely anhydrous?

Options

A. 0.1475
B. 0.2822
C. 0.8525
D. 0.4327

Answer

To find the percentage loss in water when BaCl2·2H2O becomes completely anhydrous, we can follow these steps:

### Step 1: Identify the formula and components
The compound we are dealing with is barium chloride dihydrate, which is represented as BaCl2·2H2O. This means that for every mole of BaCl2, there are 2 moles of water (H2O) associated with it.

### Step 2: Calculate the molar mass of BaCl2·2H2O
To find the percentage loss in water, we first need to calculate the molar mass of BaCl2·2H2O.

- Molar mass of Ba (Barium) = 137 g/mol
- Molar mass of Cl (Chlorine) = 35.5 g/mol (there are 2 Cl atoms)
- Molar mass of H2O (Water) = 18 g/mol (there are 2 H2O molecules)

Calculating the total molar mass:
\[ \text{Molar mass of BaCl2·2H2O} = 137 + (2 \times 35.5) + (2 \times 18) = 137 + 71 + 36 = 244 \text{ g/mol} \]

### Step 3: Calculate the mass of water in BaCl2·2H2O
Next, we need to find the mass of the water in the compound:
\[ \text{Mass of water} = 2 \times 18 = 36 \text{ g} \]

### Step 4: Calculate the mass of anhydrous BaCl2
When BaCl2·2H2O is heated, it loses the water and becomes anhydrous BaCl2. The molar mass of anhydrous BaCl2 is:
\[ \text{Molar mass of BaCl2} = 137 + (2 \times 35.5) = 137 + 71 = 208 \text{ g/mol} \]

### Step 5: Calculate the loss of mass
The loss of mass when BaCl2·2H2O becomes anhydrous BaCl2 is:
\[ \text{Loss of mass} = \text{mass of BaCl2·2H2O} - \text{mass of BaCl2} = 244 \text{ g} - 208 \text{ g} = 36 \text{ g} \]

### Step 6: Calculate the percentage loss in water
To find the percentage loss in water, we use the formula:
\[ \text{Percentage loss} = \left( \frac{\text{Loss of mass}}{\text{Initial mass}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage loss} = \left( \frac{36 \text{ g}}{244 \text{ g}} \right) \times 100 \approx 14.75\% \]

### Final Answer
The percentage loss in water when BaCl2·2H2O becomes completely anhydrous is approximately **14.75%**.

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