The percentage weight of Zn in which vitriol `[ZnSO_(4)*7H_(2)O]` is approximately equal to (at. Mass of Zn =65 , S= 32, O=16 and H=1)

Options

A. `33. 65%`
B. `32. 56%`
C. `23.65%`
D. `22.65%`

Explanation

To find the percentage weight of Zinc in vitriol (ZnSO₄·7H₂O), we will follow these steps:

### Step 1: Calculate the Molecular Weight of ZnSO₄·7H₂O
The molecular formula of vitriol is ZnSO₄·7H₂O. We need to calculate the total molecular weight by adding the atomic masses of each component.

1. **Zinc (Zn)**: Atomic mass = 65 g/mol
2. **Sulfur (S)**: Atomic mass = 32 g/mol
3. **Oxygen (O)**: There are 4 oxygen atoms in sulfate (SO₄), so:
- 4 O = 4 × 16 = 64 g/mol
4. **Water (H₂O)**: There are 7 water molecules, and the molecular weight of water is:
- H₂O = 2 × 1 (for H) + 16 (for O) = 18 g/mol
- For 7 water molecules: 7 × 18 = 126 g/mol

Now, we can sum these values to find the total molecular weight of ZnSO₄·7H₂O:

\[ \text{Molecular weight of ZnSO₄·7H₂O} = 65 + 32 + 64 + 126 = 287 \text{ g/mol} \]

### Step 2: Calculate the Weight of Zinc in ZnSO₄·7H₂O
From the formula, we know that there is 1 atom of Zinc in ZnSO₄·7H₂O, which weighs 65 g.

### Step 3: Calculate the Percentage Weight of Zinc
The percentage weight of Zinc can be calculated using the formula:

\[ \text{Percentage weight of Zn} = \left( \frac{\text{Weight of Zn}}{\text{Molecular weight of ZnSO₄·7H₂O}} \right) \times 100 \]

Substituting the values we found:

\[ \text{Percentage weight of Zn} = \left( \frac{65}{287} \right) \times 100 \]

Calculating this gives:

\[ \text{Percentage weight of Zn} \approx 22.65\% \]

### Final Answer
The percentage weight of Zinc in ZnSO₄·7H₂O is approximately **22.65%**.

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