The crystalline salt `Na_(2)SO_(4).xH_(2)O` on heating loses 55.9 % of its weight. The formula of the crystalline salt is

Options

A. `Na_(2)SO_(4).5H_(2)O`
B. `Na_(2)SO_(4).7H_(2)O`
C. `Na_(2)SO_(4).2H_(2)O`
D. `Na_(2)SO_(4).10H_(2)O`

Explanation

To find the formula of the crystalline salt \( \text{Na}_2\text{SO}_4 \cdot x \text{H}_2\text{O} \) based on the information that it loses 55.9% of its weight upon heating, we can follow these steps:

### Step 1: Determine the initial weight of the salt
Assume the initial weight of the crystalline salt \( \text{Na}_2\text{SO}_4 \cdot x \text{H}_2\text{O} \) is 100 grams.

### Step 2: Calculate the weight after heating
Since the salt loses 55.9% of its weight during heating, the weight remaining after heating is:
\[ \text{Weight after heating} = 100 \, \text{g} - 55.9 \, \text{g} = 44.1 \, \text{g} \]

### Step 3: Identify the weight of water lost
The weight of water lost during heating is:
\[ \text{Weight of water lost} = 55.9 \, \text{g} \]

### Step 4: Calculate the molar mass of anhydrous \( \text{Na}_2\text{SO}_4 \)
The molar mass of \( \text{Na}_2\text{SO}_4 \) can be calculated as follows:
- Sodium (Na): \( 2 \times 23 = 46 \, \text{g} \)
- Sulfur (S): \( 32 \, \text{g} \)
- Oxygen (O): \( 4 \times 16 = 64 \, \text{g} \)

Thus, the total molar mass of \( \text{Na}_2\text{SO}_4 \) is:
\[ 46 + 32 + 64 = 142 \, \text{g/mol} \]

### Step 5: Use the mass ratio to find the molar mass of water
We know that the remaining mass (44.1 g) corresponds to the anhydrous salt, and the mass lost (55.9 g) corresponds to the water. We can set up a proportion to find the number of moles of water:

Using the ratio:
\[ \frac{55.9 \, \text{g}}{44.1 \, \text{g}} = \frac{x \cdot 18 \, \text{g/mol}}{142 \, \text{g/mol}} \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ 55.9 \times 142 = 44.1 \times x \times 18 \]
Calculating the left side:
\[ 55.9 \times 142 = 7933.8 \]
Now, calculating the right side:
\[ 44.1 \times 18 = 793.8 \]
Setting the equation:
\[ 7933.8 = 793.8x \]
Now, solving for \( x \):
\[ x = \frac{7933.8}{793.8} \approx 10 \]

### Conclusion
The formula of the crystalline salt is:
\[ \text{Na}_2\text{SO}_4 \cdot 10 \text{H}_2\text{O} \]