`13.4 g` of a sample of unstable hydrated salt `Na_(2)SO_(4).XH_(2)O` was found to contains `6.3 g` of `H_(2) O`. The number of molecular of water of crystalistion is

Options

A. 5
B. 7
C. 2
D. 10

Explanation

To solve the problem of determining the number of molecules of water of crystallization in the hydrated salt \( \text{Na}_2\text{SO}_4 \cdot x \text{H}_2\text{O} \), we can follow these steps:

### Step 1: Calculate the mass of the anhydrous salt
The total mass of the hydrated salt is given as \( 13.4 \, \text{g} \) and the mass of water \( \text{H}_2\text{O} \) is \( 6.3 \, \text{g} \).

To find the mass of the anhydrous salt, we subtract the mass of water from the total mass:
\[ \text{Mass of anhydrous salt} = \text{Total mass} - \text{Mass of } \text{H}_2\text{O} = 13.4 \, \text{g} - 6.3 \, \text{g} = 7.1 \, \text{g} \]

### Step 2: Calculate the moles of anhydrous salt
Next, we need to calculate the number of moles of the anhydrous salt \( \text{Na}_2\text{SO}_4 \). The molar mass of \( \text{Na}_2\text{SO}_4 \) is calculated as follows:
- Sodium (Na): \( 23 \, \text{g/mol} \times 2 = 46 \, \text{g/mol} \)
- Sulfur (S): \( 32 \, \text{g/mol} \)
- Oxygen (O): \( 16 \, \text{g/mol} \times 4 = 64 \, \text{g/mol} \)

Adding these together gives:
\[ \text{Molar mass of } \text{Na}_2\text{SO}_4 = 46 + 32 + 64 = 142 \, \text{g/mol} \]

Now, we can calculate the moles of anhydrous salt:
\[ \text{Moles of anhydrous salt} = \frac{\text{Mass of anhydrous salt}}{\text{Molar mass of } \text{Na}_2\text{SO}_4} = \frac{7.1 \, \text{g}}{142 \, \text{g/mol}} \approx 0.050 \, \text{moles} \]

### Step 3: Calculate the moles of water
Now, we calculate the moles of water \( \text{H}_2\text{O} \). The molar mass of water is \( 18 \, \text{g/mol} \):
\[ \text{Moles of } \text{H}_2\text{O} = \frac{\text{Mass of } \text{H}_2\text{O}}{\text{Molar mass of } \text{H}_2\text{O}} = \frac{6.3 \, \text{g}}{18 \, \text{g/mol}} \approx 0.35 \, \text{moles} \]

### Step 4: Determine the ratio of moles of water to moles of anhydrous salt
To find the number of moles of water of crystallization per mole of anhydrous salt, we take the ratio of the moles of water to the moles of anhydrous salt:
\[ \text{Ratio} = \frac{\text{Moles of } \text{H}_2\text{O}}{\text{Moles of anhydrous salt}} = \frac{0.35 \, \text{moles}}{0.050 \, \text{moles}} = 7 \]

### Conclusion
Thus, the number of molecules of water of crystallization \( x \) in the hydrated salt \( \text{Na}_2\text{SO}_4 \cdot x \text{H}_2\text{O} \) is:
\[ \boxed{7} \]