The hydrated salt, `Na_(2)SO_(4)*nH_(2)O` undergoes `55.9%` loss in weight on heating and becomes anhydrous. The value of `n` will be:

Options

A. `5`
B. `3`
C. `7`
D. `10`

Explanation

To determine the value of \( n \) in the hydrated salt \( Na_2SO_4 \cdot nH_2O \) that undergoes a 55.9% loss in weight upon heating, we can follow these steps:

### Step 1: Understand the composition of the hydrated salt
The hydrated salt is represented as \( Na_2SO_4 \cdot nH_2O \). Here, \( Na_2SO_4 \) is sodium sulfate and \( nH_2O \) represents the water of hydration.

### Step 2: Calculate the molar mass of the components
- The molar mass of \( Na_2SO_4 \):
- Sodium (Na): \( 22.99 \, g/mol \times 2 = 45.98 \, g/mol \)
- Sulfur (S): \( 32.07 \, g/mol \)
- Oxygen (O): \( 16.00 \, g/mol \times 4 = 64.00 \, g/mol \)

Total for \( Na_2SO_4 \):
\[ 45.98 + 32.07 + 64.00 = 142.05 \, g/mol \]

- The molar mass of water \( H_2O \):
\[ 2 \times 1.01 + 16.00 = 18.02 \, g/mol \]

### Step 3: Write the total molar mass of the hydrated salt
The total molar mass of \( Na_2SO_4 \cdot nH_2O \) is:
\[ 142.05 + 18.02n \, g/mol \]

### Step 4: Calculate the mass loss upon heating
When the hydrated salt is heated, it loses water. The percentage loss in weight is given as 55.9%. This means that the mass of water lost is 55.9% of the total mass of the hydrated salt.

If we let \( M \) be the total molar mass of the hydrated salt:
\[ M = 142.05 + 18.02n \]

The mass of water lost is:
\[ \text{Mass of water lost} = 0.559 \times M \]

### Step 5: Set up the equation for the mass of water
The mass of water in the hydrated salt is \( 18.02n \). Therefore, we can set up the equation:
\[ 18.02n = 0.559 \times (142.05 + 18.02n) \]

### Step 6: Solve for \( n \)
Expanding the equation:
\[ 18.02n = 0.559 \times 142.05 + 0.559 \times 18.02n \]
\[ 18.02n - 0.559 \times 18.02n = 0.559 \times 142.05 \]
\[ (18.02 - 0.559 \times 18.02)n = 0.559 \times 142.05 \]
\[ (18.02(1 - 0.559))n = 0.559 \times 142.05 \]
Calculating \( 1 - 0.559 \):
\[ 1 - 0.559 = 0.441 \]
Now substituting:
\[ (18.02 \times 0.441)n = 0.559 \times 142.05 \]
Calculating the left side:
\[ 7.94982n = 79.41945 \]
Now solving for \( n \):
\[ n = \frac{79.41945}{7.94982} \approx 10 \]

### Conclusion
The value of \( n \) is approximately 10.