Question

The percentage weight of Zn in white vitriol [ZnSO4. 7H2O] is approximately equal to (at. mass of Zn= 65, S=32, 0=16 and H= 1)

• A. secondary phloem
• B. cork
• C. cambium
• D. all of these.

Answer 1

The Correct Option is D

To find the percentage weight of Zinc (Zn) in white vitriol, \( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} \), we first need to calculate the molar mass of the compound.

1. Calculate the atomic masses of each component in the compound:
   • Zinc (Zn): \( 65 \, \text{g/mol} \)
   • Sulfur (S): \( 32 \, \text{g/mol} \)
   • Oxygen (O): \( 16 \, \text{g/mol} \times 4 \, \text{(in } \text{SO}_4\text{)} = 64 \, \text{g/mol} \)
   • Water (H2O): \( (2 \times 1) + 16 = 18 \, \text{g/mol} \) per molecule
2. Calculate the molar mass of the water of crystallization:
   • Moles of water: 7
   • Molar mass of water: \( 7 \times 18 = 126 \, \text{g/mol} \)
3. Add the masses to find the total molar mass of \( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} \):
   • Total molar mass: \( 65 + 32 + 64 + 126 = 287 \, \text{g/mol} \)
4. Calculate the percentage of Zinc in the compound:
   • Percentage of Zn: \( \left( \frac{65}{287} \right) \times 100 \approx 22.65\% \)

Therefore, the percentage weight of Zinc in white vitriol (\( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} \)) is approximately 22.65%.