The hydrated salt `Na_2 SO_4. nH_2 O` undergoes 56% loss in mass on heating and becomes anhydrous. The value of 'n' will be :

Options

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

Answer

To find the value of 'n' in the hydrated salt Na₂SO₄·nH₂O that undergoes a 56% loss in mass upon heating, we can follow these steps:

### Step 1: Understand the Problem
We know that the hydrated salt loses 56% of its mass when heated to become anhydrous. We need to find the number of moles of water (n) in the hydrated salt.

### Step 2: Define the Masses
1. The molar mass of Na₂SO₄ (sodium sulfate) is 142 g/mol.
2. The molar mass of water (H₂O) is 18 g/mol.

### Step 3: Write the Total Mass of the Hydrated Salt
The total mass of the hydrated salt can be expressed as:
\[ \text{Total mass} = \text{mass of Na}_2\text{SO}_4 + \text{mass of water} = 142 + 18n \]

### Step 4: Calculate the Mass Loss
The mass loss upon heating is given as 56% of the total mass:
\[ \text{Mass loss} = 0.56 \times (\text{Total mass}) = 0.56 \times (142 + 18n) \]

### Step 5: Calculate the Mass of Anhydrous Salt
After losing water, the mass of the anhydrous salt (Na₂SO₄) is:
\[ \text{Mass of anhydrous salt} = \text{Total mass} - \text{Mass loss} \]
\[ = (142 + 18n) - 0.56 \times (142 + 18n) \]
\[ = (1 - 0.56)(142 + 18n) = 0.44(142 + 18n) \]

### Step 6: Set Up the Equation
The mass of the anhydrous salt is equal to the mass of Na₂SO₄, which is 142 g:
\[ 0.44(142 + 18n) = 142 \]

### Step 7: Solve for n
1. Distribute 0.44:
\[ 0.44 \times 142 + 0.44 \times 18n = 142 \]
\[ 62.48 + 7.92n = 142 \]

2. Rearrange the equation:
\[ 7.92n = 142 - 62.48 \]
\[ 7.92n = 79.52 \]

3. Divide by 7.92:
\[ n = \frac{79.52}{7.92} \]
\[ n \approx 10.04 \]

### Step 8: Conclusion
Since 'n' must be a whole number, we round 10.04 to 10. Thus, the value of 'n' is approximately 10.

### Final Answer
The value of 'n' is 10, so the hydrated salt is Na₂SO₄·10H₂O.

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